(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "Mathematics for technical schools"

ATHEMATICS 

FOR 

HNICAL SCHOOI 



WARk N A 



( ) K 



i Kh C < J I' ° CL 



MATHEMATICS 



FOR 



TECHNICAL SCHOOLS 



J. M. WARREN, B.A., 

Assistant Principal Day Schools 
Central Technical School, Toronto 



W. H. RUTHERFORD, MA., D.Paed., 

Director Department of Mathematics 
Central Technical School, Toronto 



TORONTO 

THE COPP CLARK COMPANY, LIMITED 

1921 



Copyright, Canada, J921, by The Copp Clark Company, Limited, 
Toronto, Ontario 



PREFACE 

In this book an attempt has been made to present the 
subject of Elementary Mathematics in a way suitable to 
industrial students in our technical schools. While it would 
be manifestly impossible to deal with the mathematics of all 
the industries in a book of this nature, yet we hope that 
the fundamentals as herein presented will form a basis for a 
wide range of industries. 

No doubt experts in the various departments will have 
suggestions to make as to how the book might be improved. 
We will be very glad to hear from them in this connection. 

With respect to the general plan of the work, we are 
indebted to Dr. F. W. Merchant, Director of Technical 
Education for Ontario, and Dr. A. C. McKay, Principal of 
the Toronto Technical Schools. Thanks are due in a special 
sense to Volney A. Ray, M.A., of the Department of Shopwork 
in the Central Technical School in connection with the 
chapter on " Mathematics of the Machine Shop," and to A. J. 
Stringer, M.S.A., of the Department of Architecture and 
Design in connection with the chapter on '• Application of 
the Measures to the Trades." The cuts of Quick Change 
Gears are by courtesy of the R. K. Le Blond Machine Tool 
Company, Cincinnati, and those of the Planimeters by courtesy 
of the Hughes Owens Company, Montreal. The drawings 
were made by James Hanes a former student of our school. 

June, 1921. 



CONTENTS 



Chapter Page 

i. — The Fundamental Operations of Arithmetic . . 1 

ii.— Fractions — Percentage 18 

hi. — Weights and Measures — Specific Gravity . . 37 

iv.— Square Root 50 

v. — Application of Measures to the Trades . . .54 

vi. — Algebraic Notation 71 

vii. — Simple Equations 84 

viii. — The Fundamental Operations of Algebra . . 92 

ix. — Formulas 103 

x. — Mensuration of Areas 108 

xi.— Ratio and Proportion 134 

xii. — Simultaneous Equations — Formulas (continual) . 140 

xiii.— Graphs 150 

xiv. — Mathematics of The Machine Shop .... 171 

xv. — Logarithms 228 

xvl— Mensuration of Solids ....... 241 

xvii.— Resolution into Factors 263 

xviii. — Indices and Surds 270 

xix.— Quadratic Equations 27!) 

xx.— Variation ' . . . 288 

« 

xxi. — Geometrical Progression 297 

Miscellaneous Exercises 301 

Tables— Decimal Equivalents. Weight and Specific 

Gravity, Logarithms, Amttlogartthms . . . 315 

Answers 323 

Index 335 



CHAPTER I. 

THE FUNDAMENTAL OPERATIONS OF ARITHMETIC. 

1. The Symbols of Arithmetic are 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. 
These symbols are called numbers, digits or figures. Their 
values depend on how they are written with respect to each 
other. When used separately or with commas between them 
as above they denote one, two, three, four, five, six, seven, 
eight, nine, zero. When written one after the other with no 
marks between, their values are determined by their positions. 
The established method of numeration, the Decimal System 
(from the Latin word decern, ten) is based on the number ten. 
For example 534 is read five hundred and thirty four. The 
figure 4 being in the first place counting from the right indi- 
cates 4 units, the figure 3 being in the second place from the 
right indicates ten times three units or thirty, the figure 5 
being in the third place from the right indicates one hundred 
times five units or five hundred. The following table indi- 
cates the values of the figures owing to their positions: 

CD 

a 

S co *s 

X3 G cd G 

CD co XI -*> co 

3 T3 co -*j T3 3 

S 1 M S 1 ,- I } J 

1 •§ H § •§ m ■ m £ -3 I P 

S fl a o 5 c --i 

§ B H H W H P E 

7 2 5 6-438 

in which a point called the decimal point is used to separate the 
units figure from one having one tenth the value. Thus 
7256-438 is read seven thousand', two hundred and fifty six 
and four hundred and thirty eight thousandths. The figures 
following the decimal point are read as thousandths because 

1 



-*J 


T3 


T3 


a 


o 
s-, 


c3 

CO 

S3 


a 


O 


3 


xi 


w 


H 



1. 


36 


2. 


•734 


3. 


43689 


4. 


718965 



9. 


•34 


10. 


•435 


11. 


•03 


12. 


•075 



2 MATHEMATICS FOR T?]CHXICAL SCHOOLS 

the last figure is in the thousandths place. Thus • 13 would be 
read thirteen hundredths because the last figure is in the 
hundredths place. A whole number may be written with a 
decimal point to the right of the units place. 

Exercises I. 

Write in words the following numbers: 

5. 93-4 

6. 732-45 

7. 43 124 

8. 7986-1583 

When the meanings of the figures in their relation to the 
decimal point have been fixed, the figures to the right of the 
decimal point are not read as above. 

For example 134-56 is read one hundred and thirty four 
decimal five six or more generally one hundred and thirty 
four point five six, that is the figures to the right of the decimal 
point are merely named in their order going from left to right. 

Exercises II. 

1. Read the numbers in Exercises I making use of this 
notation. 

Express the following numbers in figures: 

2. Four hundred and thirty four. 

3. Seven hundred and forty eight and twenty six hun- 
dredths. 

4. Six thousand, four hundred and eighty two and seven 
tenths. 

5. Five million, three hundred and nine thousand five 
hundred and six and one hundred and twenty five 
thousandths. 

6. Five one-thousandths. 

7. Sixty five ten-thousandths. 

8. Three hundred and twenty five one-thousandths. 

9. Four hundred and seventy eight point three four. 

10. Five thousand, three hundred and fifty point seven 
eight six. 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 3 

2. The Four Fundamental Operations. All computations 
in Arithmetic are made by means of the four operations: 

Addition or finding the sum. 
Subtraction or finding the difference. 
Multiplication or finding the product. 
Division or finding the quotient. 

3. Addition. The sign for addition is + (plus). Thus 
6+4 means that 6 and 4 are to be added. The result 
is called the sum. If we wish to add 6 ft. 4 in. and 3 ft. 2 in. 
we must add in. to in. and ft. to ft. In a similar way when 
adding numbers it is necessary to place tens under tens, 
units under units, tenths under tenths and so on. In the 
case of numbers having no decimal part this may be done by 
keeping the margin on the right-hand side in a straight line 
and, in the case of numbers having decimal parts, by keeping 
the decimal points in a vertical column. 

For example to add 9, 75, 18, 324, 9678, 27436, and also 
37-5, 124-69, -75, 0023, 346-058, 27, the arrangement is as 

follows: 

9 37-5 

75 124-69 

18 -75 

324 -0023 

9678 346 058 

27436 27- 



37540 536 0003 

Each column is added beginning at the right. The sum 
of the figures in the units column of the first case is 40, the 
is placed in the units column, and the 4 is carried and added 
to the figures in the second column since 40 units is equal to 
4 tens and units. The sum of the figures in the tens column 
with the 4 carried over is 24, the 4 is placed in the tens column 
and the 2 is carried to the hundreds column and so on. In 
a similar way beginning at the right the sum in the second 
case is found. 



1. 



7. 



10. 



13. 



MATHEMATICS FOR TECHNICAL SCHOOLS 
Exercises III. 

Copy in your work book and add the following: 



743 


2. 


1975 


3. 


1374 


1589 




4386 




9281 


642 




721 




4962 


7593 




15935 




758 


846 




420 




63 


25-72 


5. 


35-21 


6. 


328-42 


136-01 




136-35 




736-84 


23-54 




23-48 




39-43 


7-28 




78-62 




100-26 


199-71 




91-43 




702-85 


53-92 


8. 


118-64 


9. 


321-25 


16-81 




406-21 




76-84 


4-25 ' 




325-9 




1352-41 


•85 




76-84 




•13 


3-24 




231-35 




470-02 


» 
1392-6 


11. 


21985- 


12. 


2-635 


435-84 




436-54 




18-923 


936-815 




3985-216 




29-712 


72-002 




798-005 




43 • 002 


732-54 




•792 




1-986 


•006 




43-841 




868-12 


13-021 




983-521 




125-34 


4798-058 




7648-005 




9875- 1346 


$ 89-25 


14. 


si 728 -36 


15. 


S320-51 


121-63 




256-93 




192-81 


2-87 




24-87 




568-53 


13-42 




34-25 




402-96 


829-78 




176-98 




768-34 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 5 

4. Subtraction. The sign for subtraction is — (minus). 
Thus 6 — 4 means that 4 is to be subtracted from 6. The 
result is called the difference. In subtraction the numbers 
are arranged as in addition, that is units under units, tens 
under tens, and so on. For example, to subtract 872 from 
2625 the arrangement is as follows: 

2625 
872 

1753 

2 is taken from 5 leaving 3. Since 7 cannot be taken from 2, 
1 hundred or 10 tens is borrowed from 6 hundreds and 7 tens 
are then subtracted from 12 tens leaving 5. In the third 
column there are now only 5 hundreds in the upper line. 
Since 8 cannot be subtracted -from 5, 1 thousand is borrowed 
from the 2 thousands and 8 hundreds are then subtracted 
from 15 hundreds leaving 7 hundreds. The operation may be 
performed by adding to the lower line instead of subtracting 
from the upper line, thus 2 from 5 leaves 3, 7 from 12 leaves 
5, 9 from 16 leaves 7, 1 from 2 leaves 1. 

Exercises IV. 
Copy the following examples in your work book and subtract : 



1. 


7963-428. 


6. 


11-423-8-216. 


2. 


7-48-5-12. 


7. 


235-48-132-18. 


3. 


436-2-71-253. 


8. 


2-415-0034. 


4. 


$168 -45 -$29 -25. 


9. 


21-053-9-821. 


5. 


34648-239-21. 


10. 


638-215-421-006 



5. Multiplication. The sign for multiplication is X (mul- 
tiplied by). Thus 6X4 means that 6 is to be multiplied by 4. 
The number multiplied is called the multiplicand, the number 
by which it is multiplied is called the multiplier, the result 
is called the product. Before the operation of multiplication 
can be performed it is necessary to commit to memory the 
multiplication tables following: 



MATHEMATICS FOR TECHNICAL SCHOOLS 
Multiplication Tables. 



1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


2 


4 


6 


8 


10 


12 


14 


16 
24 


18 
27 


20 


3 


6 


9 


12 


15 


18 


21 


30 


4 


8 


12 


16 


20 


24 


28 


32 


36 


40 


5 


10 


15 


20 


25 


30 


35 


40 

48 


45 
54 


50 


6 


12 


18 


24 


30 


36 


42 


60 


7 


14 


21 


28 


35 


42 


49 


56 


63 


70 


8 


16 


24 


32 


40 


48 


56 


64 


72 


80 


9 


18 


27 


36 


45 


54 


63* 


72 


81 


90 


10 


20 


30 


40 


50 


60 


70 


80 
88 


90 


100 


11 


22 


33 


44 


55 


66 


77 


99 


110 


12 


24 


36 


48 


60 


72 


84 


96 


108 


120 



In the table the second column gives the products when 1 
is multiplied by 2, 2 by 2, 3 by 2 and so on to 12 by 2; the 
third column gives the products when 1 is multiplied by 3, 
2 by 3 and so on to 12 by 3. Similarly the seventh column 
gives the products when 1, 2, 3 and so on up to 12 are mul- 
tiplied by 7. 

To multiply 8345 by 7 the arrangement is as follows : 8345 

7 

58415 
5X7 is 35 that is 3 tens and 5 units. The 5 is placed in the 
units column and the 3 is carried to the tens column. 4 X 7 is 28 
and when the 3 carried over is added the result is 31 tens the 1 is 
placed in the tens column and the 3 is carried to the hundreds 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 7 

column. 3X7 is 21 and when the 3 carried over is added 
the result is 24 hundreds. The 4 is placed in the hundreds 
column and the 2 is carried to the thousands column. 8X7 is 
56 and with the 2 carried over the result is 58 thousands. 
The 8 is placed in the thousands column and the 5 in the ten 
thousands column. 

6. Powers of 10. 10X10=100 and may be written 10 2 . 
10X10X10=1000 and may be written 10 3 . 10 2 may be 
called the second power of 10, 10 3 the third power and so on, 
the figure placed to the right and above the ten being called 
the index or exponent of the power. It may be observed that 
the number of ciphers is the same as the index of the power. 

7. Multiplication by 10 and its Powers. 436X100 = 43600. 
725-26 X 10 = 7252-6 since 6 hundredths multiplied by 10 
becomes 60 hundredths or 6 tenths, 2 tenths multiplied by 10 
becomes 20 tenths or 2 units and so on. Also 43 -568X100 = 
4356.800. The rule may be stated as follows: — To multiply 
by 10 or its powers write the number with decimal point moved 
as many places to the right as the number of ciphers in the power, 
that is as many places as the index. Since 400 = 4X100 it is 
evident that the product when multiplying by 400 may be 
obtained by multiplying by 4 and then moving the decimal 
point two places to the right. 

Exercises V. 

Copy in your work book the following examples and find 
the products: 

1. 173X9. 9. 78-64X100. 17. 23 01X800. 

2. 187X7. 10. 1-475X8. 18. -00078X10 5 . 

3. 769X8. 11. 298X6. 19. -00846X9000. 

4. 34X7X4. 12. 298X60. 20. 1-475X800. 

5. 769X8X9. 13. 78-64X10 2 . 21. 236-896X10 6 . 

6. 296X10. 14. 0067X7. 22. -6345X4X10 3 . 

7. 345X100. 15. -0067X700. 23. 12-73X8X10 4 . 

8. 76-45X10. 16. 4-2905X10 3 . 24. -765X10 3 X9. 



8 MATHEMATICS FOR TECHNICAL SCHOOLS 

8. When the multiplier contains more than one digit the 
arrangement is as follows: 364X28= 364 

28 

2912 

728 

10192 
364 is multiplied by 8 as before. When multiplying by 2 
proceed as before but since the 2 is 2 tens the first figure 8 
of the partial product is placed in the tens column and so on. 
The partial products are added and the product 10192 obtained. 
When the numbers have decimal parts as 13-742X4-3 the 
arrangement is as follows : 13 • 742 

4-3 

41226 
54968 



59 • 0906 
The number of decimal places in the product is equal to the 
total number of decimal places in the two numbers multiplied. 

Exercises VI. 

Copy the following examples in your work book and find 
the products: 

1. 364X9. 16. -054X-721. 

2. 793X8. 17. 82-9X4-31X-08. 

3. 436X11. 18. -7854X-09X11-2. 

4. 731X43. 19. 3 009X721 -3X23 08. 

5. 936X72. 20.. 5-43X -034X7- 18. 

6. 119X27. 21. • 035 X -728X436. 

7. 4392X435. 22. 43 -9 X 16-8 X 002. 

8. 3854X729. 23. 143 • 5X7- 25 X -075. 

9. 9386X538. 24. 12-961X32-4X5-03. 

10. 1234X567. 25. 84-21 X 15-8 X 072. 

11. 23-54X21. 26. 46X08X-921. 

12. 734-183X36. 27. 736 X -98X4- 12. 

13. 98-43X13-2. 28. 158 X -75 X • 1625. 

14. 93 • 02 X- 75. 29. • 0625 X -04 X 025. 

15. -754X-028. 30. • 1416X3- 1416X3-5. 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 9 

9. Division. The sign for division is •*■ (divided by). 
Thus 35-7-7 means that 35 is to be divided by 7, or that it is 
required to find how many times 7 is contained in 35. The 
number to be divided is called the dividend, the number by 
which it is divided the divisor, the result of the division the 
quotient. When a number is not contained an exact number 
of times the part left over is called the remainder. Division 
may also be indicated thus 8 7 5 . 

10. Short Division. In general when the divisor is not too 
large the method of short division is used. 

Thus, 5852-r-7 = 7 /5852 
836' 

7 is contained in 58, 8 times and 2 to carry, 7 is contained in 25, 

3 times and 4 to carry, 7 is contained in 42, 6 times. When 

there is a remainder it is written over the divisor or reduced to 

decimal form: 

8/35826 8/35826-00 

or 



4478| 4478-25 

11. Division by 10 and its Powers. To divide by 10 2 the 
dividend may first be divided by 10 and the resulting quotient 
then divided by 10. Since dividing by 10 makes each figure 
equal to one-tenth its original value owing to position, it is 
evident that the result may be expressed thus: — to divide 
by 10 or its powers move the decimal point as many places to 
the left as the number of ciphers in the power of 10, that is as 
the index of the power. Since 600 = 6 X 100 if 600 is the divisor 
it is only necessary to divide by 6 and then move the decimal 
point two places to the left. Hence the rule: — To divide by 
a number ending with one or more ciphers move the decimal 
point in the dividend as many places to the left as the number 
of ciphers in the divisor and then divide by the part of the divisor 
preceding the ciphers. 



10 MATHEMATICS FOR TECHNICAL SCHOOLS 

Exercises VII. 
Copy in your work book the following examples and per- 
form the operations indicated: 



1. 


131948 -h4. 


7. 


13-25-r-lO 3 . 


2. 


2170944 -r- 12. 


8. 


W- 


3. 


12-348-r7. 


9. 


12 • 5 -r- 500. 


4. 


W- 


10. 


7659-^90. 


5. 


•00632 


11. 


153 18 -h 900, 




4 


12. 


9/280765. 


6. 


176 -MO 2 . 


13. 


7/324-94. 



12. Long Division. The method of long division is indicated 
by the following example: 

13/6942 /534 
65 

44 
39 

52 
52 

13 is contained in 69, 5 times. 13X5 is 65 which subtracted 
from 69 leaves 4. Bring down 4 the next figure of the dividend . 
13 is contained in 44, 3 times. 13X3 is 39 which subtracted 
from 44 leaves 5. Bring down 2 the next figure of the dividend. 
13 is contained in 52, 4 times. 13X4 is 52 which subtracted 
from 52 leaves no remainder. When there is a remainder it 
-may be written over the divisor or changed to a decimal as 
in short division 62563 ~ 39 = 62563 • 00 ■*■ 39 
39/62563- / 1604^ 
. 39 

235 
234 

163 
156 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 11 

or 
39/62563 • 00 /1604-17 + 
39 

235 
234 

163 
156 

70 
39 

310 
273 

37 

When the divisor or the dividend or both have decimal 
figures the position of the decimal point in the quotient may 
he obtained by paying attention to the following rules: 

1. When the number of decimal places in the dividend exceeds 
the number in the divisor, divide as if the divisor contained no 
decimals and point off a number of decimal places in the quotient 
equal to the number in the dividend minus the number in the divisor. 

2. When the number of decimal places in the dividend is 
less than the number in the divisor, annex zeros to the right of 
the dividend until a sufficient number of decimals has been 
obtained and proceed as before. 

6-79/57-20575/8425 



54 


32 


2 


885 


2 


716 




1697 




1358 




3395 




3395 



12 MATHEMATICS FOR TECHNICAL SCHOOLS 

Since there are 5 places in the dividend and 2 in the divisor, 
the number in the quotient is 5-2 or 3 and the quotient is 
therefore 8 -425. 

3430/16-807/4 
13 720 

3 087 

and the quotient is -004 since there are three decimal places 
in the dividend and none in the divisor. If it is required to 
carry the division to another decimal place add to the right 
of the decimal and then divide into 30870 thus: 

3430/ 16-8070/49 
13 720 ' 

3 0870 
and the quotient is -0049. 

The following examples show a method often used in deter- 
mining the position of the decimal point. 

Example:— Divide 433-652 by 163. 

2-660 
163/433 • 652 
326 

1076 
978 

985 
978 

72 

Explanation: — When the divisor is an integer the point 
in the quotient should be placed directly above the point 
in the dividend and the division performed as in whole 
numbers. 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 13 



.e:— Divide 27' 


4289 by 1-24. 




22- 12 


12 


4-/27 42-89/ 




24 8 




262 




248 




148 




124 



249 

248 



1 

Explanation: — When the divisor contains decimal figures move 
the point in both divisor and dividend as many places to the 
right as there are decimal places in the divisor. This is equiva- 
lent to multiplying both divisor and dividend by the same num- 
ber, 100 in the above, and does not change the quotient. 
Then place the decimal in the quotient above the position 
of the decimal point in the dividend and divide as in 
whole numbers. 

Exercises VIII. 

Find results to two decimal places: 



1. 


5462 -:- 84. 

4 


7. 


12354-406. 


2. 


1024 -f- 16. 


8. 


738-14-92-6. 


3. 


31264 -r- 46. 


9. 


1934 -43 4- 136 -3 


4. 


746215^-352. 


10. 


138-424-034. 


5. 


8344-6-21. 


11. 


128-942-7- -4327 


6. 


7342 -r- 26 -4. 


12. 


43-2198. 



41-8. 

13. Relative Importance of Signs of Operation. If only + 
and — signs occur they may be operated in any order. Thus 
12+3 — 2+9 — 6 = 16 in whatever order the signs are used. 



14 MATHEMATICS FOR TECHNICAL SCHOOLS 

If only X and ■*■ signs occur they must be operated in the 
order given 12-7-3X5 -7- 2 means that 12 is divided by 3, the 
quotient multiplied by 5 and the resulting product divided 
by 2. 

If + and — signs occur together with X and -5- signs the 
X and ■+■ signs must be used first and then the -f- and — signs 
may be used in any order. Thus 12-7-3+8X2-6-7-2+7 = 
4 + 16-3 + 7 = 24. 

If brackets are used as in 36 -t- (4+8) the part within the 
bracket is to be regarded as one quantity and the operation 
would be 36-7-12 = 3. 

Exercises IX. 

Find the values of: 

1. 16-r-8+4X2X3-16X2^-4. 

2. 60-25-^-5 + 15-100-7-4X2. 

3. 17X3+27 -7-3-40X2H-5. 

4. 864 -r- 12 -124 -=-31 +54 -T- 27. 

5. 13X9X62+44 V4- 17X22. 

6. 4963-V-7 + 144-T-72-14X9. 

7. 1728-s- (36-2X12) + (13 X 12) -s-(8-s- 2). 

14. Factors — Cancellation. The factors of the number are 
the integers (meaning whole numbers) which multiplied 
together give the number. Thus 3 and 5 are the factors of 

15 since 3X5 = 15. 

A number that has no factors but itself and unity (or 1) 
is called a prime number. If a prime number is used as a 
factor it is called a prime factor. Thus 2 and 5 are prime 
factors of 20. When the same number is a factor of two or 
more numbers it is said to be a common factor of those numbers. 
Thus 3 is a common factor of 27 and 36. By means of factors 
it is often possible to shorten the work in division. In 183 + 15 
since 3 is a factor common to 183 and 15 we can divide by 
it and then 183 -t- 15 = 61 -t- 5 = 12£. 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 15 

This method, cancellation, may be used in finding the value 
of such an expression as : 

2 1 2 
4X3X14X32 = 4X3XHX32 = 128 _ H , 

3X2X3X21 ' 3X2X3X21 9 " 9 ' 

1 1 3 

First the 3 below the line is divided into the 3 above the line 

and since 3-7-3 = 1 the 3's are cancelled by each other and 

l's are placed in their stead. Similarly 2 below the line cancels 

2 in the 4 above the line; next since 7 is a common factor of 14 

and 21 it is divided into 14 giving 2 and into 21 giving 3. When 

all common factors are cancelled the remaining numbers are 

128 
multiplied together giving —~- = 14?,. 



Exercises X. 



Find the values of: 



- 57X119X16 g 76-5X9-2X11 

* 17X12X19. ' 36-8X9X10. 

20X56X12 32-18X -006X3-4 

' 21X10X18. * 1-7X16-09X-003. 

77X100X18X14 42X-36X4-8 

25X11X49X16. " 1-2X-7X1-8. 

1200X515X70X100 -- 192X16-8X4-4 

5X35X103. ' 4X2-1X22. 

114X1728X999 10-24X7-29X36 

96X270X33. ' 1-44X9X1-8. 

99-25 + 14X7 10 2 X8-6X0625 

50-T-2X18. ' 2-5X4-3X2. 

2560-5-4 + 125X4-14x76 7-2X12-5X39 

17X27+32X40-1618. 1-3X1-2X10 2 . 

Exercises XI. 

Applied Problems. 

1 . In an electrical shop there were three motors, one weighed 
278 lb., another 380 lb., and the third 475 lb. What was the 
total weight? 



16 MATHEMATICS FOR TECHNICAL SCHOOLS 

2. Three coal sheds contained respectively 63821b., 14728 lb., 
24725 lb. How many tons in all three? 

3. Electric light wire was run around the four sides of two 
rooms. If the first room was 18 ft. long and 12 ft. wide; the 
second 20 ft. long and 13 ft. wide, what was the total length 
of wire required? (Electric lights require two wires). 

4. A reel of wire contained 6425 ft. If 3226 ft. were used 
on a certain job, how many ft. remained on the reel? 

5. A reel of wire contained 7280 ft. If 2348 ft. were used 
on one house and 1425 ft. on another, how many ft. were used 
on both? How many ft. were left on the reel? 

6. In the coal-bin at the school there were 48,720 lb. of coal 
at the beginning of the week. On Monday 11600 lb. were 
used; Tuesday 12350 lb.; Wednesday 10718 lb. On Thursday 
24600 lb. were received and 11880 lb. used. How much coal 
was used during these days? How much coal was there in the 
bin on Friday morning? 

7. A machinist sent in the following order for bolts: 15 
bolts, 3 lb. each; 21 bolts, 2 lb. each; 14 bolts, 4 lb. each; 
9 bolts, 3 lb. each; 11 bolts, 6 lb. each. What was the total 
weight of the order? 

8. A wiring job required the following labour: 3 men for 4 
hours each; 6 men for 5 hours each; 8 men for 9 hours 
each; 2 men for 15 hours each. Find the total number of 
hours on the job? 

9. A rod is 72 in. in length. How many pieces 5 in. in length 
can be cut from it? Would there be a remainder? 

10. An engine requires 90 lb. of coal per mile. How far 
could it run on 8 tons? 

11. If 4 dozen screws weigh one pound, how many case 
containing 24 screws could be filled from 30 lb. of screws'.' 

12. A train runs from Toronto to Penetang, a distance of 
101 miles, in 4 hours. What is the average rate per hour? 

13. The cost of construction of a railway from Toronto to 
Montreal, a distance of 333 miles, was $3,425,625. What 
was the average cost per mile? 

14. How many gallons of water would be discharged in an 
hour by two pipes, if one discharged 18 gallons per minute and 
the other 4 gallons more per minute? 



THE FUNDAMENTAL OPERATIONS OF ARITHMETIC 17 

15. If 18 men working 8 hours a day, can do a piece of 
work in 12 days, how many days will- it take 24 men working 
9 hours a day? 

16. If a horse-shoe weighs 8 oz., how many horse-shoes will 
36 lb. of steel produce. (1 lb. =16 oz.). 



CHAPTER II. 
FRACTIONS— PERCENTAGE. 

15. Definition. A yard measure is divided, or marked off, 
into three equal parts called feet so that: 

1 foot = one-third (f) of a yard. 

2 feet = two-thirds (f) of a yard. 

A foot rule is divided into twelve equal parts called inches 
so that: 

1 inch = Y2 of a foot. 
5 inches = T \ of a foot. 
9 inches = -fe of a foot. 
The symbols \, f, Y2, t \, T 9 -y are called fractions because 
they denote a part or fraction of something which has been 
divided. 

A fraction may then be defined as a number which denotes 
one or more of the equal parts into which some thing or unit 
has been divided. In the fraction f the number below the 
line is called the denominator since it denotes or names the 
parts into which the unit has been divided, the number above 
the line is called the numerator, since it denotes the number 
of parts taken. The numerator and the denominator are 
called the terms of the fraction. A fraction expressed in this 
notation is called a vulgar fraction. 

Since 4 = 5-4-7 a fraction may also be regarded as a case of 
indicated division. 

16. Kinds of Fractions. When the numerator is less than 
the denominator the fraction is said to be a proper fraction, 
Ex. \, f, f. When the numerator is greater than the denomi- 
nator the fraction is said to be an improper fraction, Ex. 
■f, I, \\. A combination of an integer (whole number) and a 
fraction is called a mixed number, Ex. 3|, 5&, 4|. 

18 



FR A CTIONS— PERCENTAGE 



19 



17. To change a Fraction to another Equal in Value but 
with Different Denominator. 



p 


§§ 




p 


/ /yyyy//. 




■p 





m^V/Ws. 




Wyyy 


■ 

^%% 




III 


IIP 




m 


JfH 






Fig. 1 

In the above figure, in each case, we have a square If inches 
to the side. 

In the first case the shaded portion contains three of the 
six equal parts and is one-half the whole figure so that \=%. 

In the second case the shaded portion contains eight of the 
twelve equal parts and is two-thirds of the whole figure so 
that \=h- 

In the third case the shaded portion contains fifteen of the 
twenty equal parts and is three-fourths of the whole figure so 



20 MATHEMATICS FOR TECHNICAL SCHOOLS 

that f =ift. Further, § may be obtained from § by multiplying 
numerator and denominator by the same number 3, also T \ may 
be obtained from f by multiplying numerator and denominator 
by the same number 4, and ^f may be obtained from f by multi- 
plying numerator and denominator by the same number 5. 

From these illustrations it may be inferred that a fraction 
is not changed in value when the numerator and the denominator 
are multiplied by the same number. 

If the above results are written f = \, ■& = §, -B = i it may 
be inferred that a fraction is not changed in value when the 
numerator and the denominator are divided by the same number. 

Exercises XII. 
Change the following: 

1. f to an equivalent fraction having 14 as denominator. 

2. f to an equivalent fraction having 24 as denominator. 

3. -i\ to an equivalent fraction having 50 as denominator. 

4. 4s to 7oths. 8. H to 144ths. 

5. f to 40ths. 9. -fc to 252nds. 

6. 1 to 56ths. 10. ft to 189ths. 

7. if to 108ths. 

18. Reduction to Lowest Terms. A fraction is said to be in 
its lowest terms when no number other than 1 will exactly 
divide the numerator and denominator or, in other words, when 
the numerator and denominator have no common factor. 

The fraction | is in its lowest terms because 7 and 9 have 
no common factor. 

The fraction f| is not in its lowest terms because 3 is a 
common factor of 12 and 15 and dividing numerator and 
denominator by the common factor, \l becomes f. 

To reduce a fraction to its lowest terms divide both parts by 
any common factor and continue the process until no further 
division is possible. Ex. \\ '} = \ \ = f . 

Exercises XIII. 
Reduce to lowest terms: 

1. |. 3. |1. 5. 7. H&. 9. WV 

2. A- 4. -£!£. 6. j 8. y 10. «*• 



FRACTIONS— PERCENTAGE 21 

19. To Reduce an Improper Fraction to a Mixed Number. 
Example: — Reduce x | 4 to a mixed number. 126 -H 5 gives 25 
for quotient and 1 for remainder and, as in division, may be 
written 251. Therefore J -f 6 - = 25i. Hence the rule: — Divide 
the numerator by the denominator and express as in division. 
Note — any integer may be written in the form of a fraction 
thus 25 = Y. 

To Reduce a Mixed Number to an Improper Fraction. 
Example: — Reduce 16f to an improper fraction. 

Since in 1 there are 7 sevenths, in 16 there are 16X7 or 112 
sevenths, and f is 3 sevenths, then 16f is 112+3 or 115 
sevenths, therefore 16f= L p. Hence the rule: — Multiply 
the whole number by the denominator and add the numerator 
to the product. Take this result for the numerator and the 
original denominator for the denominator. 

Exercises XIV. 

•Express the following improper fractions as whole or mixed 
numbers: 

1. f 4. V. 7. 

2. J£. 5. V-. 8. 

3. -V-. 6. H 5 - 9. 
Reduce to improper fractions: 

15. 3|. 17. 7 r \- 19. 

16. 6-J5. 18. llSaV 20. 

20. Addition and Subtraction of Fractions. When fractions 
have the same denominator they can be added by adding 
the numerators, and subtracted by subtracting the numerators. 

Exs. §+!=f. f-i-f 

When the denominators are not alike as \ and \ , they cannot 
be added without first changing to equivalent fractions having 
the same denominator. 

1 1X2 o 



w- 


10. 


W- 


13. 


HF 


if- 


11. 


25000 

1 • 


14. 


.159 
8 1 • 


w 


12. 


w- 






ions: 
121f. 


21. 


43If. 


23. 


722* 


91f. 


22 


400011. 


24. 


392J 





1 1X3 

2 — 2X3 


then, 


14.1-3X2-5 
2 1 3 ~ f, I — 6 > 


also, 


2_3_in_ i- JL 

3 5 15 la 15' 



22 MATHEMATICS FOR TECHNICAL SCHOOLS 

21. Least Common Multiple — Least Common Denominator 
of Fractions. The fractions J, f , f , can be added only when the 
denominators are alike. This may be any number of which the 
different denominators are factors, but in practice it is customary 
to take the smallest number containing the different denomi- 
nators. This number is then called the Least Common Multiple 
(L.C.M.) of the denominators because it is the least number into 
which the numbers will divide without remainder. It is also 
called the Least Common Denominator (L.C.D.) of the fractions. 

In the given case 12 is the L.C.M. of 2, 3, 4, then since 2 is 
contained in 12, 6 times, the numerator and denominator 
are multiplied by 6, so that \ = - % %, also § = T \, and f = T 9 T , 
men 2T3T4 -nTntii --- if- 

When the L.C.M. cannot be-easily determined by inspection 
the following method may be used: 

Find the least common multiple of 12, 14, 15, 16, 18, 20. 
2/12 14 15 16 18 20 



2/ 6 


7 15 


8 


9 10 


3/ 3 


7 15 


4 


9 3 



7 5 4 3 
L.C.M. -2X2X3X7X5X4X3 = 5040. 

Explanation: — Divide through by the least number which 
is a divisor of two or more of the given numbers. Continue 
this process until there is no number common to any two as a 
factor. In the third line 3 and 5 are struck out because 15 
is also in that line, and any number which is a multiple of 15 
is also a multiple of 3 and 5. The L.C.M. is obtained as 
indicated. 



Exercises XV. 






Find the values of the following: 






i. i+f 6. a+a-A. 


11. 


&2 I3 1 ^8' 


2. *+f 7. f+-^_ K 4._ ]V 


12. 


V-I+4i-2 


3-A-i 8. &+f+|+^. 


13. 


51+7^+6^ 


4- 8"T3 — A- 9. g +«°f ~f~3 2 ~~ 2- 


14. 


V-A + 7 9 . 


5. !+&-&. 10. 2i+4i+5£. 







FRACTIONS— PERCENTAGE 
Exercises XVI. 



23 



1. Four castings weigh respectively 8| lb., 5| lb., llf lb., 
and 7f lb. What is their total weight? 

2. A piece of steel on a lathe is 1 in. in diameter. In the 
first cut ^ in. is taken off, in the second cut A in., in the 
third cut -& in. Find the diameter of the finished piece. 



-*.£- 

+ H«^ 



7" 



-tf- 



Fig. 2 



3. Find the overall length for the template in Figure 2. 



r^hi- 



■ftH 



Fig. 3 



4. Find the missing dimensions in Figure 3. 

5. A drawing calls for the following divisions: 



3& in., 1\ in., 4fin., 8| in. 



Find the overall dimensions. 



-TV 



^■-H-*'^ 



;>r 



Fig. 4 
A crank-pin has the dimensions given in Figure 4. If 
in. is allowed at each end for finishing what must be the 



6. 



length of the rough forging? 




Fig. 6 



24 



MATHEMATICS FOR TECHNICAL SCHOOLS 



7. A drawing for a part of the end of a valve rod is given in 
Figure 5. Find the missing dimension. 



2»i* 




Fig. 6 



8. Find the missing dimensions, AB, CD, in Figure 6. 




9. Find the missing dimensions x, y, z, in Figure 7. 

10. Find the missing dimension in the upper part of the 
height in -Figure 7. 



FRACTIONS— PERCENTAGE 



25 



22. Multiplication of Fractions — Consider the following 
example: — A man left £ of his estate to his children, \ of this 
being left to his eldest son. What fraction of the estate did 
the eldest son receive ? 




Fig. 8 



We might represent this example by the above diagram. 
ABCD represents the whole estate. The shaded part ABEF, 
| of the whole, represents the part left to the children. One-half 
of this is taken, BEHG, to represent the eldest son's share, i.e., \ 
of £ or \ Xf. We further observe that, of the eight squares 
in the figure, the eldest son has three or § of the whole, 

• Ivl-3 
• • 2^4 — 8- 

From this illustration we may infer that in order to mul- 
tiply \ by £ we multiply the numerators for a new numerator 
and the denominators for a new denominator. Hence the 
rule: — To multiply two or more fractions together, multiply 
the numerators for a new numerator and the denominators 
for a new denominator. 

±IlUb, 4 Af — 4X7 — 2"8' 

Frequently cancellation shortens the process. 
Thus, fX4-X£=W- 

Make a drawing to illustrate that ^ X f = 1% -. 
To multiply a mixed number by an integer one of two methods 
may be used. 

Thus to multiply 25| by 7. 25 X7 = 175 

3 X i = \ = &% 



175+2i = 177i 
or, 25|X7 = ^X7 = 5 P = 177i 



26 MATHEMATICS FOR TECHNICAL SCHOOLS 

To multiply two mixed numbers together change each to an 
improper fraction and then multiply. 

Thus to multiply 2\ by 43. 2§X4§ = JX¥ = W = 10iV- 

Exercises XVII. 
Perform the operations indicated: 



1. 


1X4. 


9. 


tVX-VX^. 


17. 


9iXfVX2. 


2. 


5X|. 


10. 


1 3 V 7 V 54 y 9 

1 8 AfffAif AtJ. 


18. 


31X^X1. 


3. 


3Xif 


11. 


T6 Xgj AXJATS- 


19. 


ix-vxii. 


4. 


ly3 
2 A 8' 


12. 


4 y 5 1 V * 
T3 A -3- A T o- 


20. 


3|X2|X4|. 


5. 


4 A iff. 


13. 


25 V 75 Y64 
3 A "8" ^TO* 


21. 


13JXl^VX3i 


0. 


2 OI 8' 


14. 


15V 7 V 18 

nAnA 2T- 


22. 


1§X2|X1 T V. 


7. 


l n f 5 
4 0I 18' 


15. 


15^X4. 


23. 


"2 X3 T 9 X jg. 


8. 


3 V S V 2 

Iff A a Aj. 


16. 


6iXf 


24. 


62 X4 T 3 ^X I3V 



23. Division of Fractions. Consider Figure 8, page 25, 
regarding multiplication of fractions. 

ABCD represents the whole estate. The shaded part, 
I of the whole, represents the part left to the children. This 
is divided into two parts to represent the eldest son's share 
i.e., f -f- 2 or I -7- \ -. We observe that, of the eight squares in the 
figure, the eldest son has three or f of the whole. 



a • 2 _ 1 
4 • 1 — 8 • 



But in the previous illustration f = f X5. 

• 2 . 2 _ 3 v/ 1 
• • 4 • 1 — 4 A2- 

That is we may infer that to divide, f by \ we invert \ 
obtaining \ and then multiply f by \. 

Hence the rule: — To divide one fraction by another invert 
the divisor and proceed as in multiplication. 

Thus to divide f by f . Invert the divisor f (i.e., write it f) 
and multiply \ by f, .\ f -s- f = f Xf = if = T V 

To divide a mixed number by a fraction change the mixed 
number to an improper fraction and proceed as above. 

Thus to divide 16| by |. 16§-^ = 4^- Xf = ^ = 81f. 



FRACTIONS— PERCENTAGE 27 

51 
To reduce a complex fraction say =| to a simple fraction 
proceed as follows: 3 

7| - .23. " -TT ' -TT 2 A 23 **' 

Exercises XVIII. 

Find the results of the following: 

1. f-5-5. 7. i^-J. 12. 3* -5- 5* X^+^. 

2. f-f-2. 8. 125***. 13- 12-H-T-3f+4f-h4 T V 
3-A-4. i_Xj 14. 5*Xtt-3|-*-2f 

4. A-3. y 'fxr 15 ljX2U-9ft 

5. tV"^"?' 10- 8X3 "I" 6- 031^1^ 

S. tt+f 11. 4fXl|-2iX T V 16- 3|X5*-f-i-f+li-J-3*. 

24. Decimal Fractions. The values of the figures in any 
number depend upon their position with reference to the 
decimal point. 

Thus, -2 = 2 tenths = T % 

• 25 = 2 tenths + 5 hundredths = T % + to = -nnr 
•342 = 3 tenths +4 hundredths +2 thousandths 

— 3 4-_4 4- 2 - 342 

~io 1 Toll 1 1000 — T00"ff' 

In all such cases the decimal parts may be written as fractions 
with some power of 10 as denominator, and are therefore 
called decimal fractions. 

25. To change a Decimal Fraction to an equivalent Vulgar 
Fraction. It is evident that it is only necessary to write the 
decimal, after removing the point, as numerator and 1 followed 
by as many 0's as there are figures in the decimal as denominator. 

Exercises XIX. 

Change to equivalent fractions in their lowest terms: 
1. -43. 6. -004. 



2. 


•04. 


7. 


•705. 


3. 


•752. 


8. 


• 1234. 


4. 


•7134. 


9. 


•016. 


5. 


•502. 


10. 


•000155. 



28 MATHEMATICS FOR TECHNICAL SCHOOLS 

26. To change a Vulgar Fraction to its equivalent Decimal 
Fraction. Example: — Change | to its equivalent decimal. 
1, . q _8 /l-000 _ 19 , 

Example: — Change H to its equivalent decimal. H = 11 + 16 = 
16/1 1-0000 / -6875 
9 6 

140 
128 

120 
112 

80 
80 

It is evident that, to change a fraction to its equivalent 
decimal fraction, it is only necessary to perform the division 
indicated after 0's have been placed to the right of the 
decimal point. 

Exercises XX. 

Change the following fractions to their equivalent decimals: 

1. b 4. H. 7. H- 10. Hh 

2. f 5. f. 8. «. 11. n- 

3 3 A 124 Q 24 10 1 27 

27. Repeating Decimals. Example: — Change | to its equiva- 
lent decimal. ^ = 1-7-3 = 3 /1-000 

•333-1- 
The division in this case would never end. £ therefore 
produces what is known as a repeating decimal. This is 
expressed by placing a period above the figure 3 .*.^= -3. 
Example: — Change f to its equivalent decimal. 
1 = 5^6 = 6/5-0000 
•8333 + 
In this case the decimal does not begin to repeat until the 
second figure and is therefore called a mixed repeating decimal. 

.'. | =-83. 



FRACTIONS— PERCENTAGE 29 

The denominators in Exercises XX contain only 2's or 5's 
or 2's and 5's as their factors. The fractions can be changed 
into fractions having some power of 10 as denominators and 
therefore give terminating decimals. All fractions such as 
s, f, etc., having some factor other than 2 or 5 in the de- 
nominator, when expressed in their lowest terms, cannot be 
changed into fractions having some power of 10 as denomi- 
nator and therefore give repeating or mixed repeating 
decimals. 

• Exercises XXI. 

'Change the following to their equivalent decimals: 
is 2-i- 3 1 - 4 2 ^ * fi 2 7 * 

1. 9 . •£• 12- <->• T- ** TF* °- T¥- °- TT- '-T3"' 

28. To change Repeating and Mixed Repeating Decimals to 
their equivalent Fractions. 

Example: — Change -24 to its equivalent fraction 

• 24 = • 242424 

100 times • 24 = 24 • 242424 

1 times -21= -242424 

Subtracting, 99 times -24 = 24 
. 24 = i 4 

That is to change a repeating decimal to its equivalent fraction 
write the decimal, after removing the point, as numerator and 
as denominator as many 9's as there are figures in the repeating 
part. 

Example : — Change • 34 to its equivalent fraction 

•34= -34444 

100 times -34 = 34-444 

10 times -34 = 3-444 

Subtracting, 90 times -34 = 31 

. o\ _ 3 1 



30 MATHEMATICS FOR TECHNICAL SCHOOLS 

That is to change a mixed repeating decimal to its equivalent 
fraction subtract the part which does not repeat from the whole 
giving the numerator, and for denominator take as many 9's 
as there are figures in the repeating part followed by as many 
O's as there are figures which do not repeat. 

Exercises XXII. 
Express as fractions in their lowest terms: 

1. -5 6. -369 11. 2-5306 

2. -36 7. 3-253 12.* -04726* 

3. -36 8. -2516 13. -0036 

4. -153 9. -i42857 14. -0426 

5. -369 10. 2-76 

29. Percentage. The term "percent." usually written 
%, is an abbreviation of the Latin "per centum" which means 
by the hundred. Five percent. (5%) would be yf^ of the 
quantity named. Percent, may be changed to a decimal 
fraction. 

Thus, 62% = rffr=-62. 
37-5% = fftf= : 375. 
A decimal fraction of a quantity may be expressed as 
percent. 

Thus, -7 = T W = 70% 
•89 = T Vxr = 89% 
• 375 = Htf =37-5% 

That is the decimal fraction may be changed to percent, by 
moving the decimal point two places to the right. Also any 
fraction may be changed to percent, by changing it to its equivalent 
decimal fraction, and then moving the decimal point two places 
to the right. 



FRACTIONS— PERCENTAGE 31 

Exercises XXIII. 
1. In the following table supply the missing quantities: 



% 


Decimal 
Fraction 


Vulgar 
Fraction 


% 


Decimal 
Fraction 


Vulgar 
Fraction 


% 


Decimal 
Fraction 


Vulgar 
Fraction 


1 






16f 






100 








•02 


1 


37| 


•25 
•5 


1 

3 


200 






2| 




1-75 










2* 


6| 




3-86 






•10 


1 
8 






3 

4 






l-i'V 




•9 


' 


350 







2. Find 25% of 16, of 8, of 90, of 240. 

3. 5 is what % of 10? of 20? of 40? 

4. 8 is what % of 16? of 40? of 24? 

5. What % of f is 2\1 27| of 600? 

6. 20% of what number is 3? 7? 14? 17? 

7. 68 is 15% less than what number? 

8. 98 is 40% more than what number? 



32 MATHEMATICS FOR TECHNICAL SCHOOLS 

9. A gas bill was 25% higher last month than this. If it id 
$6.46 this month how much was it last month? 

10. How much water must be added to a 5% solution of a 
certain liquid to make a 2% solution? (original solution 20 
gallons) . 

30. Short Methods. In practical work a large number of 
decimal places is not needed. In all measurements the accuracy 
depends upon the instruments, the methods used, and the thing 
measured. It is only necessary that the error is small compared 
with the quantity measured; a fraction of an inch in a dimension 
of several feet would probably not make much difference. 

In measuring to -001 inches it is not necessary to carry the 
work to say -00001 inches. In any case of multiplication or 
division it is only necessary to carry the result to one decimal 
place more than the measurement. Thus if a measurement 
of 7-265 inches is multiplied by 3- 1416 it is only necessary to 
carry the work to four places of decimals, care being taken to 
allow for numbers carried over from the fifth place. 

Other short methods of multiplication and division may 
be used. 

To multiply by 5, 50, 500, etc., add 0, 00, 000, etc., to 
the right of the number and divide by 2. Why? 

To multiply by 25, 250, etc., add 00, 000 to the right of the 
number and divide by 4. \\^hy? 

To multiply by 125, add 000 to the right of the number 
and divide by 8. Why? 

To multiply by 33*, 16|, 12$, 8|, 6J. Add 00 to the 
right of the number and divide by 3, 6, 8, 12, 16. Why? 

By using the reverse process division by 33£, 16f, 12£, 
125, etc., may be performed. Thus to divide by 33^ multiply 
by 3 and divide by 100 or mark off two decimal places. Why? 

To multiply a number ending in \ such as \Z\ by itself. 
Multiply the number plus 1 by itself and add \ to the product. 

Thus 13|X13i = 14Xl3 + l. 



FRACTIONS— PERCENTAGE 33 

To multiply a number ending in 5 by itself, multiply the 
number to the left of 5 by a number one greater than itself and 
place 25 to the right of the number. Thus, 75X75, 7X8 = 56, 
and the result is 5625. 

Exercises XXIV. 
Applied Problems. 

1. From 2000 lb. of iron bars each weighing 80 lb. § is cut 
up for bolts, £ for shafts and the remainder for studs. How 
many bars are used for the different articles? 

2. At 2^c. a pound, what will be the cost of 108 castings 
each weighing 29 lb.? 

3. An automobile runs at the average rate of 10| miles an 
hour. How long will it take to go from Toronto to London, 
a distance of 116 miles? 

4. A | in. steel bar weighs 1-914 lb. per foot. What will be 
the cost of 5000 ft. of f in. steel bars if it cost $1.75 per 100 lb.? 

5. Which is cheaper, and by how much, to have a 36^c. 
an hour man take 12| hr. on a job or to have a 48|c. an hour 
man who can do the job in 9| hr.? 

6. The weight of a foot of ^ in. steel bar is 1-06 lb. Find 
the weight of a 20 ft. bar. 

7. At 42^ c. an hr. what will be the pay for 21 \ days of 8 
hours each? 

8. If 2\ bundles of shingles are used on 82| sq. ft. of roof, 
how many bundles will be used on 325 sq. ft. of roof? 

9. How many pieces 5| in. long can be cut from a rod 27 
ft. long? 

10. A person spending ^, § and \ of his money has SI 19 
left; how much had he at first? 

11. If T 4 T of a house be worth $1969.92, what is the value 
of ^ of the house? 

12. Three men own a house worth $6250; one owns T 3 „ of it; 
the second £ of it; what is the value of the third's share? 

13. A man having 271 J acres of land, sold £ to one man 
and | to another; what was the value of the remainder at 
$323 • 68 an acre? 

14. I want to mix up a pound of solder to consist of '4 parts 
zinc, 2 parts tin and 1 part lead; what fraction of a pound of 
each metal must I have? 



34 MATHEMATICS FOR TECHNICAL SCHOOLS 

15. An apprentice who is drilling and tapping a cylinder for 
| in. studs, tries a f in. drill, but the tap binds, so he decides 
to use a drill -fa in. larger; what size drill will he use? 

16. An 8 ft. bar of steel is cut up into 16 in. lengths; what 
fraction of the whole bar is one of the pieces? 

17. The time cards for a certain piece of work show 2 hours 
and 15 minutes lathe work, 4 hours and 10 minutes milling, 
2 hours and 20 minutes bench work; what is the total number 
of hours charged to the job? 

18. A gallon is about -^ of a cubic ft. If a cubic foot of 
water weighs 62? lb., how much does a gallon of water weigh? 

19. What is the cost of a casting weighing 432^ lb. at 6|c. 
a pound? 

20. How many steel pins to finish 1| in. long can be cut 
from an 8 ft. rod if we allow ^ in. to each pin for cutting off 
and finishing? 

21. A machinist whose rate is 67-5 cents per hour puts in a 
full day of 8 hours and also 3 hours overtime. If he is paid 
"time and a half" for overtime, how much should he be paid 
altogether? 

22. If an alloy is -67 copper and -33 zinc, how many pounds 
of each metal would there be in a casting weighing 82 lb.? 

23. A can do a piece of work in 25 days; B can do it in 30 
days; C can do it in 35 days. In what time will they do it, 
all working together? 

24. A man earns $280 in 2\ months. If he spends in 4 
months what he earns in 3 months, how much will he save in a 
year? 

25. From a farm of 125 T 3 7 acres there were sold at one time 
27-63 acres and at another 34| acres. How many acres 
remained? 

26. From an oil tank containing 375-087 gallons there 
leaked out each day 2f gallons. How many gallons remained 
in the tank at the end of 25 days? 

27. If the weight of a brass casting is approximately fifteen 
and a half times that of its white pine pattern, what will be 
the weight of a casting if the pattern weighs 15 oz.? 

28. Since the shrinkage of brass castings is about \ in. in 
10 in., what length would you make the pattern for a brass 
collar which is required to be 6 in. long? 



FRACTIONS— PERCENTAGE 35 

29. How long will it take a drill making 134 revolutions per 
minute (R.P.M.), at the rate of -012 in. per revolution, to 
drill a hole 1^ in. deep? 

30. A piece of wrought-iron 2-69 in. thick is to have two 
H in. holes drilled through it. If the drill makes 112 R.P.M., 
what must be the feed to drill each hole in two minutes? 
(The feed of a drill is the number of revolutions necessary to 
cause the drill to descend 1 in.). 

31. In drilling a bed plate a drill makes 67 R.P.M., and is 
being fed to the work at the rate of -015 in. per revolution, 
how deep will the hole be at the end of 4| minutes? 

32. What will be the R.P.M. of a drill used for drilling a 
lathe spindle 30-24 in. long, the feed being -015 in. per revo- 
lution, and the time given to the job being 21 minutes? 

33. What must be the R.P.M. of a drill, feeding at the 
rate of -015 in. per revolution, to drill a hole 2\ in. deep in a 
casting in 2 minutes? 

34. A casting is to have a number of holes drilled in it 2\ 
in. deep with a high-speed drill making 260 R.P.M. What 
must be the feed to drill each hole in f of a minute? 

35. A man who owns f of a claim sold -6 of his share for 
$2000. What decimal part of the claim does he still own and 
what is the claim worth? 

36. An engine rated at 1250 horse-power, is found to be 
y o efficient. How many horse-power are available for 

driving the machinery? How many are lost? 

37. A board was cut into two pieces, one 8f in. and the 
other 5^ in. long. If ^ in. be allowed for waste in cutting, 
what was the length of the board? 

38. A locomotive has a piston displacement of 12656 cu« 
in. If the clearance space is 6-5% of the piston displacement, 
what is the clearance space? 

39. A merchant bought 15 carloads of apples of 212 barrels 
each, 3 bushels in each barrel at 90c a bushel. He paid for 
them in cloth at 25c. a yard. How many rolls of 477 yd. each 
did he give? 

40. A carload of pig-iron weighs 90,000 lb. If 11$% of 
this is used at once in the foundry, how much is left? 

41. The diameter of two holes is 3| in. and the distance 
between the sides of the holes is 3f in. What is the distance 
from the outside of one hole to the outside of the other? 



36 MATHEMATICS FOR TECHNICAL SCHOOLS 

42. From a steel bar 27 f in. long were cut the following 
pieces: — one 1\ in., one 6| in., one 3f in. long. If the length 
of the bar was then 8| in., what was the amount of waste in 
cutting? 

43. A man, buying a house and lot, paid $2200 for the lot 
and 62|% more than that for the house. What did both 
cost him? 

44. A man invested $16,400 as follows: — 25% in an auto- 
mobile, 37^% in bank stock, and the remainder in an addition 
to his house. How much did he invest in each? 

45. An electrician has a reel of 300 ft. of copper wire. He 
used at various times 50| ft., 32j ft., 109f ft. How much 
wire was left? What percent, was left? 

46. If § of the shell of a stationary boiler is considered as the 
heating surface, how many square feet of heating surface are 
there in a boiler containing 98^ sq. ft.? 

47. A pump pumps 3-38 gallons to each stroke and the 
pump makes 51-2 strokes per minute. How many gallons of 
water will it pump per hour? 



CHAPTER III. 
WEIGHTS AND MEASURES— SPECIFIC GRAVITY. 

31. Linear Measure. Linear Measure is used in measuring 
lines and distance. 

The fundamental unit of English Linear Measure is the 
yard. It is the distance between two marks on a bronze 
bar in the Royal Exchange, London, England. 

Table. 
12 inches (in.) = 1 foot (ft.). 
3 ft. =1 yard (yd.). 

5^ yd. = 1 rod. 

320 rods = 1 mile. 

Inches are commonly denoted by two strokes above the 
figure. Feet are denoted by one stroke. Thus 6 in. is written 
6" and 6 ft. is written 6'. 

32. Surveyor's Measure. Surveyor's Measure is used in 
measuring land. 

Table. 
7 -92 in. = 1 link (li.). 
100 li. = 1 chain (ch.). 
80 ch. = 1 mile. 
1 ch. =22 yd. =66 ft. 

The chain in this table is known as Gunter's chain. It is 
the one in general use for country surveys. 

Engineers frequently use a chain, or steel tape, 100 ft. long. 
The feet are usually divided into tenths instead of into inches. 

33. Nautical Measure. 

Table. 

6 ft. =1 fathom. 

120 fathoms = 1 cable. 

6080 ft. = 1 nautical mile = 1 • 151 statute miles. 

1 knot = a sailing rate of one nautical mile per 

hour. 

37 



38 MATHEMATICS FOR TECHNICAL SCHOOLS 

Exercises XXV. 

1. How many yards in a mile? 

2. How many feet in a mile? 

3. One inch is what decimal of a yard? 

4. One rod is what decimal of a mile? 

5. Reduce 18 yd., 2 ft., 9 in. to inches. 

6. Reduce 3 mi., 30 rods, 1^ yd. to feet. 

7. Express 1 link as a decimal of a mile. 

8. Express 1 in. as the decimal of a chain. 

9. Change 4 chains, 15 links to links. 

10. Change 26 yd., 1 ft., 2 in. to chains. 

11. Change 4356 li. to feet. 

12. Change 25 rods, 3 yd., 2 ft. to chains. 

13. The world's record (Dec. 1919) for a destroyer was 
45 • 5 knots. What is this in statute miles? 

34. Metric Linear Measure. Metric is the adjective form 
of the word metre which is a French word meaning "measure." 
The earth's quadrant (one fourth of the circumference) was 
measured by French engineers in 1799. One ten-millionth of 
this length was taken as the length of the metre. 

Table. 

10 millimetres (mm.) = 1 centimetre (cm.) 
10 cm. = 1 decimetre (dm.) 

10 dm. = 1 metre (m.) 

10 m. =1 decametre (Dm.) 

10 Dm. = 1 hectometre (Hm.) 

10 Hm. = 1 kilometre (Km.) 

It may be seen that the prefixes have definite meanings: 

milli = xoViT' centi = y^, deci = T x 7 , deca = 10, hecto = 100, 

kilo = 1000. 

35. Comparison of English and Metric Linear Measurements. 

1 in. =2-5399 cm. (2-54 cm. approx.). 
lcm. =-3937 in. 

lmile =1-60935 Km. (1-61 Km. approx.). 
1 Km. =-621 miles. 

1 m. =39-3707 in. (39-37 in. approx.). 
Make calculations to test the accuracy of the above table. 



WEIGHTS AND MEASURES— SPECIFIC GRAVITY 



39 



Exercises XXVI. 

1. Measure the perimeter of the room with both metre stick 
and yard stick. Make drawings to scale in your laboratory 
book. Change the result in the English system to the Metric 
system and compare. 

2. Do the same as in 1 for the door, table, etc. 

3. Write all the measurements in the Metric system in 
terms of the metre. 

4. Fill in the omitted entries in the following: 



Unit 


Equivalent 




Inches 


Feet 


1 cm. 






1 dm. 






1 m. 






1 Dm. 






1 Hm. 






1 Km. 







5. A piece of steel bar is laid off to a length of 438 cm. 
Find this length in feet and inches. 

6. The thickness of a steel plate is f ". Find the thickness 
in cm. and dm. 

7. A speed of 200 ft. per second is how many Km. per 
second? 




40 MATHEMATICS FOR TECHNICAL SCHOOLS 

8. When a body falls freely from rest it increases in speed 
each second 32-2 ft. per second. Express this in cm. per 
second each second. 

9. An express train is travelling at the rate of 50 miles per 
hr. Express this in Km. per minute. 

10. Find the difference in cm. between the lengths of two 
steel rods, one of which is 4-8' long and the other 4-8" long. 

Square Measure. In measuring areas or surfaces, the 
inch, foot, yard, etc., can no longer be used. 
It is necessary to use the square inch, the 
square foot, the square yard, etc. 

By a square inch is meant a surface one 
inch long and one inch wide. 

Thus in measuring surfaces two dimen- 
sions, length and breadth, are used. 

Table. 
144 square inches (sq. in.) = 1 square foot (sq. ft.). 
9 sq. ft. =1 square yard (sq. yd.). 

30j sq. yd. =1 square rod (sq. rod). 

160 sq. rods = 1 acre. 

10 sq. chains = 1 acre. 

640 acres =1 square mile (sq. mi.). 

Make drawings to scale in your laboratory book and illus- 
trate the truth of the first three lines in the above table. 

37. Metric Square Measure. 

Table. 
100 square mm. (sq. mm.) = 1 square cm. (sq. cm.) 
100 sq. cm. = 1 square dm. (sq. dm.) 

100 sq. dm. = 1 square m. (sq. m.) 

100 sq. m. =1 square Dm. (sq. Dm.) 

100 sq. Dm. = 1 square Hm. (sq. Hm.) 

100 sq. Hm. = 1 square Km. (sq. Km.) 

Make drawings to scale in your laboratory book and illus- 
trate the truth of each line in the above table. 



WEIGHTS AND MEASURES— SPECIFIC GRAVITY 41 

38. Comparison of English and Metric Square Measure. 

Table. 

1 sq. in. =6-4516 sq. cm. 

1 sq. cm. = • 155 sq. in. 

1 sq. ft. =-0929 sq. m. 

lsq. m. =10-764sq. ft. 

1 sq. yd. = -8361 sq. m. 

1 sq. m. =1- 196 sq. yd. 
Make calculations to test the accuracy of the above table. 

Exercises XXVII. 

1. Find the area of the floor of your classroom in square 
metres and also in square feet. Make drawings to scale in 
your laboratory book. Change the area in square metres to 
square yards and compare. 

2. Find the area of a page of your laboratory book in sq. 
in. and also in sq. cm. Test as in preceding question. 

3. Perform similar experiments by measuring the school- 
yard, the door, table, the teacher's desk, etc. 

4. Change one acre to sq. yd. 

5. Express 4 sq. rods, 25 sq. yd., 7 sq. ft., in sq. ft. 

6. Express 5 sq. rods, 8 sq. yd., 5 sq. ft., as the decimal of 
an acre. 

7. Express 5 sq. yd., 3 sq. ft., 18 sq. in., as sq. in. 

8. Express 4 sq. ft., 85 sq. in., as the decimal of a sq. yd. 

9. A square field measures 20 rods to a side. Find its area 
in acres. 

10. A steel plate in the form of a rectangle is 18^" long by 
6j" wide. Find the area in sq. ft. 

11. A number-plate on an automobile is 21" long by b\" 
wide. Find area in sq. ft. 

12. A rectangular garden 2\ chains wide contains f of an 
acre. How many feet long is it? 

13. How many sq. ft. of glass are there in a box containing 
72 panes each 12" by 16"? 

14. How many sq. yd. are there in the walls of a room 15' 
6" long, 12' wide, and 9' 4" high? 

15. A rectangular piece of land measures 1200 links by 180 
links. What is its area in acres? 



42 



MATHEMATICS FOR TECHNICAL SCHOOLS 



16. How many bricks 8 in. long and 4 in. wide will pave a 
yard 116' long and 46' wide? 

17. Find the cost of laying a concrete walk 400 yd. long 
and 4 ft. 8 in. wide at 60c. a sq. yd. 

18. Find the cost of painting both sides of a tight board 
fence 80' long, 5' 3" wide at 7c. a sq. yd. 

19. How many boards each 12' long and 10" wide will 
be required to build a fence 60 yd. long and 4 ft. high? 

20. How many sq. ft. of tin will be necessary to line the inside 
of an open box whose external measurements are 4' long, 3' 
8" wide and 2' 10" deep, if the material in the box is 2" thick 
and 10% is allowed for cutting and joining the tin? 

39. Cubic Measure. In the measurement of surfaces in the 

preceding sections two measure- 
ments, length and breadth, were 
used. The areas resulting were ex- 
pressed in square inches, square 
feet, etc. 

If it is required to measure the 
volume of solids, the dimensions, 
length and breadth must be taken 
into account and in addition 
another dimension — thickness. 

By a cubic inch is meant the 
volume of a cube, 1 inch on each 
edge, Figure 10. 

Volumes of solids are measured in cubic inches, cubic feet, 
cubic yards, etc. Table. 

1728 cu. in. = 1 cu. ft. 
27 cu. ft. = 1 cu. yd. 
128 cu. ft. = 1 cord (8'X4'X4'). 

Make drawings to scale in your laboratory book and illus- 
trate the truth of each line in the above table. 

40. Metric. Cubic Measure. 

Table. 
1000 cubic millimetres (c.mm.) = 1 cubic centimetre (c.c.) 
1000 c.c. = 1 cubic decimetre (c.dm.) 

1000 c.dm. = 1 cubic metre (cm.) 

1000 cm. = 1 cubic decametre (c.Dm.) 

1000 c.Dm. = 1 cubic hectometre (c.Hm.) 




Fig. 10 



WEIGHTS AND MEASURES— SPECIFIC GRAVITY 43 

41. Comparison of English and Metric Cubic Measure. 

Table. 
1 cu. in. =16-387064 c.c. =16-387 c.c. (approx.) 
1. c.c. =. -06102 cu. in = -061 cu. in. (approx.) 
1 cu.ft. = -02831 cm. = 028 cm. (approx.) 
1 cm. =35-3163 cu. ft. =35-316 cu. ft. (approx.) 
Make calculations to test the accuracy of the above relations. 
Exercises XXVIII. 

1. Find the volume of the top of the laboratory table in cu. 
ft. and in cm. Make drawings in your laboratory book. 
Change from one system to the other and compare. 

2. Find the volumes of the various rectangular models in 
the laboratory, in cu. in. and also in c.c. Make drawings in 
your laboratory book. Change from one system to the other 
and compare. 

3. Change 8 cu. yd., 9 cu. ft., to cubic feet. 

4. Change 3 cu. yd., 2 cu. ft., 8 cu. in., to cubic inches. 

5. A rectangular vessel is 15" long, 6£" wide and 4|" deep, 
inside measurements. Find its volume in cubic centimetres. 

6. A gravel bed whose surface has an area of 2 acres, con- 
tains gravel to a depth of 10". How many miles of road 12' 
wide can be covered with the gravel if it be spread to a uniform 
depth of 7"? 

7. The outside measurements of a cubical box, with a lid, 
are 3' 4" long, 2' 8" wide and 1' 10" deep. If the box is made of 
1" material, how many cu. ft. of material are there in the box? 
How many cubic metres will it hold? 

8. A cubical cistern, without a lid, 4' 4" long, 4' 4" wide and 
6' 8" deep, outside measurements, is made of plank 2" thick. 
How many cu. ft. of material are there in the box? How many 
cu. ft. of water will it hold? 

9. A pile of wood 10' long, 4' wide and 6' high was sold for 
$20.00. What was the price per cord? 

10. A wood-yard 20' long and 18' wide is filled with cord 
wood to a height of 6'. What is the wood worth at $8.50 a 
cord? 

11. If 1 cu. yd. of earth make a load, how many loads will 
be removed in excavating for a foundation 4' deep, 36' 3" 
long, and 24' wide? 



44 MATHEMATICS FOR TECHNICAL SCHOOLS 

12. The end of a rectangular bar of iron is a square f " to 
the side. How many c.c. are there in 4' of the bar? 

13. In excavating a tunnel 374,166 cu. ft. of earth were 
removed. If the length of the tunnel was 492 ft. and the 
width 39 ft., what was the depth? 

14. In making a tender for some excavating a contractor 
notes that the excavation is in the shape of a rectangle 11' 
wide, 86' long at the top and has a depth of 8'. What will it 
cost him to excavate it at 40c. a cu. yd.? What must he bid 
to make a profit of 15%? 

15. Rain falling uniformly for 5 hours on a roof, whose 
dimensions are 30' by 15', fills a tank 6' 3" by 3' by 2' 6". Find 
the depth of the rainfall per. hour. 

16. The ice on a pond whose area is £ of an acre is 10" 
thick. How many cu. ft. of ice may be removed? 

42. Measures of Weight. The fundamental unit of English 
weight is the pound. There are the pound Avoirdupois and 
the pound Troy. 

The pound, Avoirdupois, is equal to the weight of 7000 
grains (plump grains of wheat) and is used for all ordinary 
purposes of weighing. The pound, Troy, is equal to 5760 
grains and is used in weighing gold, silver and precious stones. 

Table — Avoirdupois Weight. 
16 drams =1 ounce (oz.). 
16 oz. = 1 pound (lb.) =7000 grains. 
100 lb. =1 hundredweight (cwt.). 
20 cwt. = 1 ton. 
2240 lb. = 1 long ton. 

Table— Troy Weight. 
24 grains = 1 penny weight (dwt.) 
20 dwt. = 1 oz. 
12 oz. =1 lb. =5760 grains. 

43. Metric System of Weights. The fundamental unit of 
metric weight is the kilogram which is the weight of 1 litre, 
equal in volume to 1 cubic decimetre, of distilled water under 
fixed conditions of temperature and pressure. 



WEIGHTS AND MEASURES— SPECIFIC GRAVITY 45 

Table. 

10 milligrams = 1 centigram (eg.) 

10 eg. = 1 decigram (dg.) 

10 dg. = 1 gram (g.) 

10 g =1 decagram (Dg.) 

10 Dg. = 1 hectogram (Hg.) 

10 Hg. = 1 kilogram (Kg.) 

44. Comparison of English and Metric Systems of Weights. 

Table. 
1 gram =15-432 grains. 
1 ounce = 28 -35 grams. 
1 pound (avoirdupois) =453*6 grams. 

= -4536 kilograms. 
1 kilogram =2-2046 pounds. 
1 metric ton = 1000 kilograms. 
- =2204-6 pounds. 
Knowing any one of the above relations test the accuracy 
of the others. 

45. Measures of Capacity. The fundamental unit of 
capacity in the English system is the gallon, which contains 10 
pounds of distilled water under fixed conditions of temperature 
and pressure. 

46. Liquid Measure — used in measuring liquids. 

Table. 

4 gills = 1 pint (pt.) 

2 pt. =1 quart (qt.) 

4 qt. =1 gallon (gal.) 

47. Dry Measure — used in measuring grains, vegetables, etc. 

Table. 
2 pints = 1 quart (qt.) 
4 qt. =1 gallon (gal.) 
2 gal. = 1 peck (pk.) 
4 pk. = 1 bushel (bu.) 



46 MATHEMATICS FOR TECHNICAL SCHOOLS 

48. Metric System. The fundamental unit of measure- 
ment is the litre and is equal in volume to one cubic decimetre. 

Table. 
10 millilitres = 1 centilitre (cl.) 
10 cl. = 1 decilitre (dl.) 

10 dl. = 1 litre (1.) 

10 1. =1 decalitre (Dl.) 

10 Dl. = 1 hectolitre (HI.) 

10 HI. = 1 kilolitre (Kl.) 

= 1 cu. metre 

49. Comparison of Capacity Tables with Cubic Measure. 

1 litre =61-024 cu. in. (approx.) 

= •22 gal. 
1 gal. =4-54 1. 
1 cu. ft. = 28 -38 litres. 

= 6-2321 gal. 
277-274 cu. in. = 1 gal. 231 cu. in. = 1 gal. (American). 

50. Specific Gravity. The specific gravity (sp. gr.) of a 
substance is its weight as compared with the weight of an equal 
volume of pure water. 

Since the weight of a fixed volume of water is known we 
can find the weight of an equal volume of any substance if we 
know the specific gravity. 

Example: — Find the weight of 8 cu. ft. of steel if its sp. gr. 
is 7.8. 

Solution: — 1 cu. ft. water weighs 62-321 lb. 

1 cu. ft. steel weighs 62-321X7-8 lb. 
8 cu. ft. steel weighs 62-321X7-8X8 lb.= 
3888-83 lb. 

Exercises XXIX. 

1. How much space will be filled by 14 tons of wrought- 
iron (sp. gr. 7-7)? 

2. Find the average sp. gr. of a piece of brick construction 
weighing 114 lb. per cu. ft. 



WEIGHTS AND MEASURES— SPECIFIC GRAVITY 47 

3. If 13 litres of milk weigh 13-39 kilograms, what is the 
sp. gr. of milk? 

4. A tunnel 625 yd. long having a cross-section of 64 sq. 
yd. is excavated through rock of sp. gr. 2-7. Find the weight 
of rock removed. 

5. If 3 litres of alcohol weigh 2-37 kilograms, what is the 
sp. gr. of alcohol? 

51. Measure of Time: 

Table. 

60 seconds (") = 1 minute (1'). 
60 minutes = 1 hour. 
24 hours = 1 day. 

7 days = 1 week. 

365 days, 5 hours, 48 minutes, 48 seconds = 1 year. 
As the calendar year of 365 days is nearly 6 hoars less than 
the above, correction is made as follows: — Every year whose 
number is divisible by 4 is a leap year and contains 366 days, 
the other years containing 365 days, except that the century 
years are leap years only when the number of the year is 
divisible by 400. 

The year is divided into 12 months: — January (Jan.), 
February (Feb.), March, April, May, June, July, August 
(Aug.), September (Sept.), October (Oct.), November (Nov.), 
December (Dec). 

" Thirty days hath September, April, June and November." 
The other months, except February, have 31 days each. 
February has 29 days in leap years and 28 days in all other 
years. 

Exercises XXX. 

1. Compute the actual number of days from Sept. 23, 1919, 
to April 6, 1920. 

2. A note bearing interest from March 8, 1899, was paid on 
July 5, 1900. Compute the interest period. 

3. Reduce to the lowest denomination named: — 4 weeks, 
3 days, 15 hr. 23 min. 

4. How many hours between 10 A.M. Jan. 1, 1920, and 
6 P.M. March 3, 1920. 



48 MATHEMATICS FOR TECHNICAL SCHOOLS 

52. Miscellaneous Measures : 

Counting Tables. 
12 things =1 dozen (doz.). 
12 doz. = 1 gross. 
12 gross = 1 great gross. 
20 units = 1 score. 

Stationers' Tables. 
24 sheets = 1 quire. 
20 quires = 1 ream. 

3 reams = 1 bundle. 

5 bundles = 1 bale. 

Exercises XXXI. 

1 Calculate the volumes of a number of the rectangular 
solid models in the laboratory and estimate their weights in 
both systems. Change from one system to the other and check. 

2. Fill in the omitted entries in the following: 



Quantity 


Volume 


Weight 




cu. in. 


c.c. 


wt. in lb. 


wt. 


in Kg. 


2 pints water 










3 qt. water 










1 cu. ft. water 










1 gal. water 


277-274 




10 




10 c.c. water 




10 






1. Kl. water 











WEIGHTS AND MEASURES— SPECIFIC GRAVITY 49 

3. A rectangular tank is 2-5 m. long, 1-4 m. wide, and 
•98 dm. deep. Find its capacity in litres. Find the weight 
of water it will hold in grams. 

4. The thickness of a steel plate is f ". If the plate has an 
area of 400 sq. dm., find its volume in cu. in. and its weight in 
lb. if 1 cu. in. of steel weighs -283 lb. 

5. A block of granite weighs 2\ tons. Find, its weight in 
kilograms. 

6. Find the weight in grams of the air in a room 16' X 10' 
and 9' high, if the air is -00128 times as heavy as water. 

7. Find the number of litres in a rectangular tank 8'X 
6'6"X4'3". 

8. How many gallons of water are contained in a tank 6 
metres long, 3-4 metres wide, and 2-7 metres deep? 

9. A concrete watering trough is Z\' wide, 8' long and 2' 
deep outside while inside the basin is 2' 10" wide, 7' 4" long 
and 1' 6" deep. What is its weight if a cu. ft. of concrete 
weighs 145 lb.? If the concrete was mixed in the proportion 
of 1 cement, 2 sand, 3 stone, and 1| cu. yd. dry material makes 
1 cu. yd. concrete, how many bags of cement were used. 
(1 bag = l cu. ft.)? 



CHAPTER IV. 
SQUARE ROOT. 

53. The Square of a Number is the product obtained by 
multiplying the number by itself. Thus the square of 5 = 
5X5 = 25. 

The square root of a given number is that number whose 
square is the given number. Thus the square root of 25 is 
5 because 5X5 = 25. 

Square root is indicated by prefixing the symbol y/ to the 
given number. Thus \A>4 denotes the square root of 64. 

When a number is small the square root may be found by 
inspection or by means of the factors of the number. Thus 
1225 = 5 X5X7 X7 = 5 2 X7 2 so that V1225 =V(5 2 X 7 2 ) 
= 5X7 = 35. 

The following general method may be used for finding the 
square root. To find the square root of 1326-4164. 

1326-4J64 /36-42 
9 

66 426 

396 

724 3041 

2896 

7282 145 64 

145 64 

Explanation: — Beginning at the decimal point, separate the 
number into groups of two figures each, counting both to the 
right and the left. Find the greatest square in the left-hand 
group and write its square root as the first figure of the root. 

In the example, 9 is the greatest square in 13, and 3 is the 

first figure in the root. 

60 



SQUARE ROOT 51 

Subtract the square from the left-hand group and to the 
remainder bring down the next period to the right, thus forming 
a new dividend. 

In the example, 9 is subtracted from 13 and along with the 
remainder 4 the next group 26 is brought down, giving 426 
as the new dividend. Divide the new dividend, with its 
right-hand figure omitted, by twice the part of the root already 
obtained and annex the result to both the root and the divisor. 

Multiply the complete divisor by the last figure of the 
root obtained, subtract, and bring down the next group to 
form a new dividend as before. 

In the example the 3 in the root is doubled giving 6, 6 is 
now divided into 42 giving 6, and this figure is placed to the 
right of the 6 already in the divisor and also as the second 
figure of the root. Although 6 divided into 42 gives 7, if this 
result is taken the result 67X7 gives a quantity too great to 
subtract from 426, so that 6 must be taken instead. Proceed 
in this manner until all the groups are used. 

For every group to the right of the decimal point there must 
be a decimal figure in the root. 

When the number is not an exact square the root may be 
obtained to any number of decimal places. 

Exercises XXXII. 

Find the square root (correct to four decimal places) of: 
1. 2025. 2. 39601. 3. 15129. 4. 106929. 5. 1369. 

6. 3. 7. 12- 186. 8. 143-2041. 9. -5432. 10. -06285. 

11. Find the length in yards of the side of a square 10 acre 
field. 

12. A square pipe has an area of 136-0752 sq. in. What is 
the length of its side? 

13. An outlet on a heating system is 4' 4" wide and 18" 
high. A pipe leading from it must have the same area and 
must be square. Find the size of the square pipe. 

14. Would it be cheaper to build the square pipe or one of 
the same dimensions as the outlet? Why? 



52 



MATHEMATICS FOR TECHNICAL SCHOOLS 



15. A steel plate is rectangular in shape, 18"X14". Find 
the side of a square plate of the same area. 

54. One of the most Valuable Practical Uses of Square Root 
is in finding the third side of a right-angled triangle, when 
two of its sides are given. 

In the adjoining figure ABC 
is a right-angled triangle with 
the sides AB and BC 3 in. 
and 4 in. respectively (to 
scale) . 

Squares are described on 
the three sides and divided 
into smaller squares as in- 
dicated. If we make tests 
with dividers we will find 
that the small squares are 
equal throughout the figure. 
We will also notice that the 
number of small squares in 
the square on AC is equal 

to the total of the number of small squares in the squares 

on AB and BC. 

From this experiment we derive: — In a right-angled triangle 

the square on the side opposite the right angle {hypotenuse) is 

equal to the sum of the squares on the other two sides. 

Exercises XXXIII. 

1. Find the distance from corner to corner of a square 
piece of tin which contains 100 sq. in. 

2. A room is 40' X 28'. Find the length of a diagonal. 

3. If the above room is 16' high, find the distance from 
any corner to the diagonal corner of the ceiling. 

4. A baseball diamond is in the form of a square 90' to the 
side. Find the distance from "first" to "third." 

5. A boy was flying a kite with a string 650' long. If the 
distance, from where the boy was standing, to a point directly 
under the kite was 450 ft. how high was the kite? 



- 



Fig. 11 



SQUARE ROOT 53 

6. A tree broke in such a way that the top struck the ground 
30' from the base of the tree. What was the height of the 
tree, the broken part being 60 ft. long? 

7. A ladder, 42' long, placed with its foot 24' from a wall, 
reached within 2' of the top. How near the wall must the 
foot of the ladder be brought in order that it may reach the 
top? 



CHAPTER V. • 

APPLICATION OF MEASURES TO THE TRADES. 

55. Stone Work. In stone work it is difficult to get any- 
fixed method of estimating the cost. One job will have a 
set of conditions which do not exist in another, hence the 
contractor will make an allowance in one that he would not 
regard as necessary in another. There are two kinds of stone 
work, rubble or rough and ashlar or squared. In rubble 
work the toise is the common unit of measurement. This is 
used both in estimating the amount of stone required and in 
the cost of the work. 

Table. 

10 tons rubble = 1 toise (approx.) 
1 toise — in wall = 162 cu. ft. (approx.) 
1 toise — measured loose = 216 cu. ft. (approx.) 
In ashlar work the unit for estimating either the amount of 
stone necessary, or the cost of laying, is the cubic foot. The 
labour for dressing the stone is figured by the square foot. The 
minimum thickness of stone work for facing is 4 in. increasing 
in thickness as requirements demand. 

Exercises XXXIV. 

1. How many toise of rubble will be required for the founda- 
tion of a house 40' 0" X 32' 0". the stone work being 5' 0" 
high and 18" thick? 

2. A cellar is 23' 6" wide by 35' 8" long and 6' 6" high. If 
the wall is 16" thick and has two openings each 3' 3" X 2' 3", 
find the number of toise of stone required. 

3. The basement walls for a house 26' 0" wide and 38' 0" 
long are to have 6 windows each 3' 0" X 2' 0". The walls are 
to be 7' 0" high and 18" thick, (a) Find the cost at $20.00 
a toise if the actual volume be estimated and 5% be allowed for 
extra work on openings. (6) Find the cost at $18.00 a toise if 
corners be doubled and only 50% of the openings be deducted. 

54 



APPLICATION OF MEASURES TO THE TRADES 55 

4. A foundation wall for a building 28' 0" X 40' 0" is to be 
7' 0" high and 1' 6" thick. There are to be 4 openings, two 
3' 0" X 2' 6" and two 3' 0" X 5' 0". Concrete is to be used 
in the construction and is to be mixed in the following pro- 
portions: — 1 cu. ft. (1 bag) of cement, 2\ eu. ft. sand and 5 
cu. ft. broken stone. If \\ cu. yd. of dry material will make 
1 cu. yd. of concrete, find the number of cu. ft. of cement, of 
sand, and of broken stone. 

5. A building 24' 6" wide, 36' 0" long and 20' 0" high, above 
the foundation, is to be of stone with walls 16" thick. The 
foundation, 6' 0" high, 16" thick, is to be concrete and to have 
6 windows 1' 10" X 3' 4". If a cu. yd. of concrete requires 25 
cu. ft. of stone, 12 cu. ft. of sand, and 4 cu. ft. of cement, find 
the number of cu. ft. of each in the foundation. In the walls 
of the house there are to be 8 windows 2' 0" X 5' 0", 3 windows 
3' 6" X 5' 0" and 3 doors 3' 6" X 7' 0". How many cu. ft. of 
stone will be necessary? 

56. Brick Work. There is the same lack of uniformity in 
methods of estimating cost in brick work as in stone work. In 
measuring up the cost of the work some contractors make no 
deduction for openings less than 2 ft. square. Usually, how- 
ever, the exact volume of the brick work is estimated and, in 
fixing the cost, allowance is made for extra labour and material 
for arches, cuttings, etc. 

Since bricks are of varying size no fixed rule for the volume 

of laid brick can be given. If we consider an ordinary stock 

brick as 8f " X 2|" X 4" and add a |" joint to thickness, length 

and width we get 9" X 2|" X 4|" or approximately 9" X 3" 

X 4|". The number of bricks for 1 cu. ft. of masonry would 

then be 9^fiJ " 14 «- 

Table— (Based on above calculation). 
Per cubic foot, 15 bricks. 

Superficial foot of 9" wall, 11 bricks. 
Superficial foot of 13" wall, 16£ bricks. 
Superficial foot of 18" wall, 22 bricks. 

The labour and material for brick work are usually estimated 
by the 1000 brick, if in a straight wall. 



56 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Exercises XXXV. 

1. Make drawings to scale, in your laboratory book, of 
bricks of different sizes. Allowing a §" joint calculate the 
number of bricks that will be required for a wall 20' 0" long, 
8' 0" high, and 18" thick. 

2. A house is to have 27' 0" frontage, 30' 0" in depth, and 
20' 0" in height above foundation. It is to have 8 windows 
4' 6" X 5' 6" and 4 doors 4' 3" X 7' 0". The wall is to be 9" 
thick; allowing 15 bricks to the cu. ft., how many bricks will 
be required? 

3. If the wall in the preceding question is 13" thick, find the 
number of bricks. 

4. What will it cost to lay the brick in each of the two 
preceding questions if a bricklayer lays an average of 700 a 
day and received 90c. an hour for an eight hour day? 

5. The walls of a building 40' 0" wide and 100' 0" long are 
to be 18' 0" high. There are 4 doors 8' 0" X 8' 0", 4 doors 3' 
3" X 7' 0", 30 windows 4' 0" X 5' 0". Making use of the table 
for superficial area find the number of bricks required, if the 
wall is 13" thick? 

6. Reckoning 15 bricks per cu. ft., find the cost at $30 a 
thousand for the walls of a building 30' 0" wide, 50' 0" long 
and 24' 0" high with the following specifications: — the lower 
14' 0" is to have a wall 18" thick and is to have 4 doors 2' 
10" X 6' 10" and 5 windows 3' 0" X 7' 0"; the upper 10' 0" 
is to have a wall 13" thick, and is to have 6 windows 3' 0" X 
5' 0". 

57. Lumber. The common unit of measurement in lumber 

is the board foot. 

It is a piece of lumber 
1 ft. long, 1 ft. wide, and 
1 in. thick. 

If we take a board 
12 ft. long, 12 in. wide, 
FlG - n and 1 in. thick, we 

readily see that it will contain 12 board feet. 




APPLICATION OF MEASURES TO THE TRADES 57 

This might have been obtained as follows: — length (in 
feet) X width (in feet) X thickness (in inches), thus 12 X 1 
X 1 = 12. 

This rule is applicable for finding the board feet of all kinds 
of lumber. Example : — Find the number of board feet of lum- 
ber in a floor joist 2" X 10", 18' 0" long. 

Solution: — Number of board feet = length (in ft.) X width 
(in ft.) X thickness (in in.) = 18X^X2 = 30. 

Lumber is billed in different ways,' (1) per thousand (M) 
board feet, (2) per thousand (M) sq. ft., (3) per foot run. 

Speaking generally we may say that, in dealing with 
material 1" thick and up, the board foot is the unit, although 
special sizes up to 2" X 3" are frequently charged as per foot 
run. Below 1" in thickness material is reckoned in sq. ft., 
except "trim" which is sold as per foot run. 

The following data for estimating the S S [3 

amount of allowance for dressing and Fiq 
working the tongue in flooring is furnished 

by one of the large lumber companies of Toronto: 

1§" wide, f " thick, add 50% 
\\" wide, I* thick, add 33% 
2" wide, I" thick, add Zl\% 
2" wide, |" thick, add 25% 
2\" wide, |" thick, add 33|% 

Example: — Find the cost of flooring a room 20' X 10' 
with No. 1 red oak flooring, \\" X f ", at $160 per M 
sq. ft. 

Solution: — Area of floor = 20 X 10 sq. ft. 

Lumber required = itnrX 20 X 10 sq. ft. 

Cost = i£S X 20 X 10 X tVW = $48.00. 



58 MATHEMATICS FOR TECHNICAL SCHOOLS 

The following is a sample bill of lumber: 



125 
400 
130 
130 
310 

500 

2000 

2000 

14 

70 

100 

100 

5 



ft. lineal, lf"X6", Pine D4S. 
ft. lineal, 1" XI", Pine Rgh. 
ft. lineal, f" X10", Pine D4S. 
ft. lineal, f " X2£", Bed Mldg 
ft. lineal, f * Xlf", Fir Picture 

Mldg 

ft. B.M., 1" No. 1 H. D1S.... 

ft. Strip 6" H. Decking 

ft. Strip | " Spruce Fig 

pieces, 2"X 4' , X12 / , H. Szd.. . 
pieces, 2" X 10" X 10', No. 1 H. 

Szd 

ft. lineal, 2" X 2", H. Rgh 

ft. lineal, 3" X2", H. Rgh 

pieces, 2|"X5f"X10 / , Oak Sill. 



125 
400 
130 
130 

310 

500 

2000 

2000 

112 

1167 

100 

50 

50 



Price 



1 

8, 
3. 

3 

62. 
66 
68 
63. 

65. 

2. 

65. 



25 
00 
00 
00 

00 
00 
00 
00 
00 

00 
00 
00 
75 



Amount 



$ 9.06 

4.00 

10.40 

3.90 

9.30 

31.00 

132.00 

136.00 

7.06 

75.86 
2.00 
3.25 

37.50 



Total. 



461.33 



D4S — dressed on four sides. 
Rgh. — rough. 
Mldg. — moulding. 
Fig. — flooring. 



Szd. — sized. 

No. 1— No. 1 (best quality). 

H. — hemlock. 

Lin. — per foot run. 



Exercises XXXVI. 

1. Take measurements of a number of pieces of lumber 
obtained from the woodworking shop. Make drawings in 
your laboratory book and estimate the board feet in each. 

2. Measure the top of a laboratory table, the top of the 
teacher's desk, etc. Make drawings in your laboratory book 
and estimate the board feet in each. 

3. Take measurements of the floor of your classroom and 
make a drawing in your laboratory book. Find the cost of 
flooring with birch 2\" wide and §" thick at $140 per thousand 
square feet. 



APPLICATION OF MEASURES TO THE TRADES 



59 



4. By means of a drawing in your laboratory book, show the 
number of board feet in a cubic foot. 

5. Find the number of cu. ft. in a stick of timber 
6" X 8" X 18' 0". Change to board feet and check by 
rule. 

6. An oak stick is 8" X 8" X 30' 0". Find its volume 
by cubic measure. Change to board feet and check by 
rule. 

7. A lot 60' 0" frontage and 120' 0" in depth is to be enclosed 
on two sides and an end by a tight board fence 6' 0" high. 
The posts are to be placed 6' 0" apart and to cost 40c. each; 
there are to be two string pieces 2" X 4" from post to 
post on which to nail the boards; the boards are to be 
1" thick. If lumber is worth $56 per M, find the total cost 
of same. 

8. What will it cost at $52 per M to cover the floor of a 
barn 32' 0" X 42' 0" with 2" square plank? 

9. A room is 12' 0" wide and 16' 0" long. Find the cost, at 
$175 per thousand square feet, of laying a No. 1 red oak floor, 
the material being \%" wide and f " thick. 

10. Complete the following bill of lumber: 





Feet 


Price 


Amount 


500 ft. lineal, f "X3f", Pulley stile 

500 ft. lineal, i"X3f ", Lining 




$4.00 
2.50 
2.50 
1.00 
9.50 
7.25 
2.00 
9.50 




500 ft. lineal, 7"X3l", Lining 




500 ft. lineal, Parting stop 




150 ft. lineal, 2"X6", Sash sill 




125 ft. lineal, lf"X6", Pine D4S 

400 ft. lineal, \" X6", Backing 




200 ft. lineal, 2" X6", Door jamb 






Total. 









Note — Prices are for 100 ft. lineal. 



60 MATHEMATICS FOR TECHNICAL SCHOOLS 

11. Complete the following bill of lumber: 



Feet 



Price 



44 pieces, 2"X12"- 
10 pieces, 8" X 14"- 
105 pieces, 2" X 4"- 
15 pieces, 2" X 4 r - 
32 pieces, 2" X 4"- 
17 pieces, 2" X 6"- 
23 pieces, 2" X 6"- 
9 pieces, 2" X 8"- 



20' 0' 
16' 0' 



long, 
long, 



10' 0" long, 

12' 0" long, 

8' 0" long, 

16' 0" long, 



12' 0' 
14' 0' 



long, 
long, 



Red Pine. 
Red Pine. 
Hem. Szd 
Hem. Szd 
Hem. Szd 
Com. Pine 
Hem. Szd. 
Hem. Szd. 



S84.00 
86.00 
63.00 
63.00 
62.00 
86.00 
65.00 
66.00 



Total. 



12. Complete the following bill of lumber: 



Price 



12 pieces, 1"X7"-16' 0" long, Pine D4S . 
2 pieces, 6"X6"-16' 0" long, Pine D4S.. 

310 ft. lineal, 8", Fir base 

680ft. lineal, f "X2f ", Fir base D4S 

46 pieces, | " X 5|"-14' 0" long, Door jamb 

sanded 

xlOOft. lineal, If "X3f ", Pine D4S 

x9 pieces, If "X5f "-10' 0"long, Pine D4S. 

5 pieces, 2f " X5f "-12' 0" long, Oak sill . . 

5 pieces, 2f "X5f"-10' 0"long, Oak sill. . 

2 pieces, 2f "X5f"- 14' 0" long, Oak sill. . 

65 ft. lineal, 3", Crown moulding 

xl20ft. lineal, l"X9f ", Clear PineD4S. . . 
xl25 f t. lineal, f " X 5f ", Clear PineD4S. . . 
xl25 ft. lineal, |"X3f ", Clear PineD4S.. . 



S 6.00 

85.00 

8.75 

5.50 

8.00 

123.00 

163.00 

.75 

.75 

.75 

3.75 

190.00 

160.00 

150.00 



Total. 



x Dressed out of material even inch above. 



APPLICATION OF MEASURES TO THE TRADES 61 

58. Roofs, Rafters, Pitch. 




In the above section of an ordinary gable roof, some of the 
terms used in connection with roofs are indicated. The span 
of a roof is the same as the width of the building. The run is 
one-half the span, and the rise is the vertical distance from the 
top of the plate to the top of the ridge. The pitch of a rafter 
is given by dividing the number of feet in the rise by the 
number of feet in the span. Thus if the rise is 6 ft. and the 
span 12 ft. the roof would have a one-half pitch. The rafter 
length is the distance from the outside corner of the plate to 
the centre of the ridge. The heel is the distance from the 
outside corner of the plate to the end of the rafter. The length 
of the heel would have to be added to the rafter length if the 
above method were used for the construction of the eaves. 



62 



MATHEMATICS FOR TECHNICAL SCHOOLS 



The accompanying figures illustrate the method of finding the 
lengths of the different rafters in a Hip or Cottage roof. Figure 




Fig. 15 



15 shows a plan of the roof, Figure 16 a right side elevation, 
Figure 17 a plane at plate level. 




In order to find the length of a hip rafter it would first 
be necessary to find the length of HK in Figure 17. Using 



APPLICATION OF MEASURES TO THE TRADES 



63 



this length and the perpendicular distance from H to the 
ridge the length of the hip rafter may be found as in Figure 18. 




Fig. 17 



To find the length of the jack rafter we observe in Figure 
P7 that, if the rafters be 16" on centre, MN would also be 16 ". 




Fig. 18 




Also since the roof has a \ pitch, the perpendicular distance 
from the hip to N would also be 16", hence Figure 19. 
If the roof has other than a \ pitch, similar triangles would 
give the lengths of the jack rafters. 



64 



MATHEMATICS FOR TECHNICAL SCHOOLS 



59. Roofing — Shingles. Shingles for roofing are estimated 
as being 16" long and averaging 4" wide. They are put 
up in bundles of 250 each, four bundles making a square of 
shingles. 

The unit in measuring for roofing is the square. A square 
contains 100 sq. ft. If shingles are laid 4" to the weather, 
each shingle would on an average cover an area of 16 sq. in. 
This would give for 100 sq. ft. -i^p or 900 shingles. In this 
result, however, no allowance has been made for waste in 
cutting or for defective shingles. 

The following table has been found useful in practice 
(Kidder's Pocket Book): 



Inches to the 


Area Covered by 


Number to Cover 


Weather 


1000 Shingles 


a Square 


4 


100 sq. ft. 


1000 


4J 


110 sq. ft. 


910 


4| 


120 sq. ft. 


833 


5 


133 sq. ft. 


752 


5£ 


145 sq. ft. 


690 


6 


156 sq. ft. 


637 



60. Roofing — Slate. Slate for roofing is also measured by 
the square (100 sq. ft.). In estimating either the amount 
required or the cost of laying, eaves, hips, valleys, etc., are 
measured extra — 1 ft. wide by the whole length. The sizes 
of slates range from 9" X 7" to 24" X 14". "Each slate 
should lap the slate in the second row below, 3 inches", 
Kidder. 

The gauge of a slate is the portion exposed to the weather, 
which should be one-half of the remainder obtained by sub- 
tracting 3 in. from the length of the slate. 






APPLICATION OF MEASURES TO THE TRADES 65 

The following table is taken from Kidder's Pocket Book: 



Size of Slates 


Inches Exposed 


Number to a 


in Inches 


to Weather 


Square 


14X24 


10i 


98 


12X24 


10§ 


115 


12X22 


9| 


126 


11X22 


9* 


138 


12X20 


8| 


142 


10X20 


8| 


170 


12X18 


n 


160 


10X18 


71 

' 2 


192 


9X18 


n 


214 


12X16 


6| 


185 


10X16 


6| 


222 


9X16 


6^ 


247 


8X16 


6| 


277 


10X14 


5* 


262 


8X14 


5| 


328 



Exercises XXXVII. 

Note. — In working the following problems take the actual 
quantity of lumber used, not allowing for waste due to having 
to buy stock lengths of material. In case of fractional inches 
take the inch above in each separate piece. 

Lumber is cut in lengths of 10' 0", 12' 0", 14' 0", 16' 0", 
18' 0", and will be charged on that basis. 

1. Find the number of shingles for a square of roof for 
each line in the table if no allowance be made for waste. 

2. A shed 9' 0" wide and 18' 0" long is to have a "lean to" 
roof, \ pitch. If the rafters are 2" X 4" at 16" centre and have 
a 12" heel, find their cost at $52 per M. If the roof extends 
12" on each end, find the cost of covering with 1" square 
sheeting at $56 per M. Find the cost of shingling the above 
with shingles laid 4|" to the weather, if material and labour 
cost $14 a square of shingles. 

3. A garage 10' 0" wide and 16' 0" long is to have a gable 
roof, \ pitch. The rafters are 2" X 4" at 2' 0" centre and have 
a 15" heel. The rafter ties are 2" X 4" X 10' 0". Find the 
cost at $50 per M. If the roof extends 10" on the ends and 




66 MATHEMATICS FOR TECHNICAL SCHOOLS 

6" more on each end be allowed for waste, find the cost of 
covering with 1" square sheeting, at $48 per M. 

Find the cost of shingling the above with shingles laid 
4?" to the weather if material and labour cost S13..50 per square 
of shingles. 

4. A stable 15' 0" wide and 20' 0" long is to have a gable 
roof, \ pitch. The rafters are 2" X 4" at 20" centre and have 

an 18" heel, the ridge board 
being 1" X 6". The roof is 
supported at every second 
rafter by a brace 2" X 4" 
(see Figure 20) and collar 
ties 2" X 6" X 15' 0". Find 
the cost of lumber at $52 
per M. If the extension on 
the ends is 12", find the cost 
of sheeting with 6" tongued 
and grooved lumber at $55 per M sq. ft., allowing 10% for 
the tongue and groove and 6" on each end for waste. 

Find the cost of shingling the above roof with shingles laid 
5" to the weather if material and labour cost $13 per square 
of shingles. 

5. A house 25' 0" wide and 32' 0" long is to have a gable 
roof, f pitch. The rafters are 2" X 6", 16" on centre, with an 
18" heel. The roof is supported by braces 2" X 4", 4' 0" on 
centre, 6' 0" long, and tied with ceiling rafters 2" X 6"X25'0". 
Find the cost of the above lumber at $55 per M. 

If the extension on the ends be 12", find the cost of covering 
with |" X 6" tongued and grooved sheeting at $66 per M sq. 
ft., allowing 10% for the tongue and groove and 8" on each end 
for waste. 

Find the cost of roofing the above with slate, the gauge 
being 8^", if material and labour cost $30 a square. 

6. A building 20' 0" wide and 28' 0" long is to have a hip 
roof, \ pitch, the ridge being 1" X 8-" X 8' 0" long. The 
hip rafters are 2" X 6", the jack rafters 2" X 6", at 16" 
centre. Ceiling joist 5' 0" from plate level, 2" X 6"; act as 
ties. Find the cost of the above lumber at $56 per M. 

Note. — In estimating the amount of material in rafters, find 
the length of a common rafter and multiply by the number 
of rafters on both sides. 



APPLICATION OF MEASURES TO THE TRADES 67 

7. A building 22' 0" wide and 32' 0" long is to have a hip 
roof, | pitch, the ridge being 1" X 8" X 10' 0". The hip 
rafters are 2" X 6", the jack rafters 2" X 6", at 16" centre; 
ceiling joist 2" X 6" X 22' 0" act as ties. There are also 14 
braces 2" X 4", at 16" centre, from centre of ceiling joist to 
centre of common rafters. Find the cost of lumber at $55 
per M. 

61. Lathing and Plastering. In lathing and plastering the 
square yard is the unit of measurement. 

Standard laths are 4 ft. long, 1\ in. wide, and are laid \ in. 
apart. They are usually put up in bundles of 50. It requires 
approximately 18 laths to cover a square yard. In estimating 
both the amount and the cost of lathing and plastering, the 
percentage of the openings deducted from the total area will 
depend upon the job. In small openings no deduction will be 
made; in medium openings about 40% or 50%; in very large 
openings from 75% to 90%. 

Exercises XXXVIII. 

1. What will it cost to lath and plaster the walls and ceiling 
of the following rooms at 70c. per square yard? (a) 17' 0" 
X 13' 0" X 9' 0" high, with a door 3' 0" X 7' 0", and 3 windows 
each 3' 3" X 5' 0". Deduct 40% of the openings. (b) 20' 
0" X 18' 0" X 11' 0" high, with 2 doors 3' 0" X 9' 0", and 4 
windows 3' 0" X 5' 0". Deduct 50% of the openings, (c) 
40' 0" X 22' 0" X 15' 0" high, with 2 doors 5' 6" X 8'0", and 
window space 15' 0" X 9' 0". Deduct 90% of the openings. 

2. A room is 16' 0" wide, 18' 0" long, and 10' 0" high. There 
are 2 doors 4' 6" X 7' 0", 3 windows 3' 8" X 4' 9", the sills 
being 2' 3" from the floor. Find (a) the cost of lathing 
and plastering at 90c. a sq. yd., deducting 50% of the openings; 
(b) the cost of laying and finishing a |" X 2" No. 1 red oak 
floor at 32c. per sq. ft.; (c) the cost of paneling the walls to a 
height of 4' 0" at 80c. a sq. ft. (Note — -plastering is carried to 
floor behind paneling.) 

3. Find the cost of lathing and plastering a room 27' 0" 
wide, 30' 0" long, and 12' 0" high at 80c. a sq. yd., if there are 
2 doors 3' 6" wide and 7' 0" high, and 6 windows 3' 4" wide and 
6' 0" high. Deduct 50% of the openings and also deduct 12c. 
a sq. yd. for the area paneled (see No. 4), on account of the 
finishing coat being unnecessary. 



68 MATHEMATICS FOR TECHNICAL SCHOOLS 

4. Find the cost of paneling the walls in the preceding 
question to a height of 4' 6" if the sills of the windows be 2' 
6" from the floor, at 85c. a sq. ft. 

5. A hall is 50' 0" wide, 90' 0" long, and 20' 0* high. There 
are 4 doors 5' 6" X 10' 0", 2 windows 5' 0" X 11' 0", 7 windows 
5' 0" X 8' 3". Find the cost of lathing and plastering at 75c. 
a sq. yd., deducting 80% of the openings. 

62. Decorating and Painting. Wall paper is put up in 
rolls, the number of yards in the roll and the width of the 
paper varying. The kinds chiefly in use are: 

(1) Paper 18" wide and in single rolls 8 yd. in length, or 
double rolls 16 yd. in length. 

(2) Paper 21" wide and in rolls 12 yd. in length. 

(3) Paper 30" wide and in rolls 5 yd. in length; frequently 
put up in 15 yd. rolls. 

To estimate for the walls. 

(1) Find the perimeter of the room, less the width of doors 
and windows. 

(2) Find the number of strips required by dividing the result 
in (1) by the width of the paper. 

(3) Find the number of strips that can be cut from a roll by 
dividing the length of the roll by the height to be papered. 

(4) Find the number of rolls by dividing the number of 
strips required for the room by the number of strips in a roll. 

To estimate for the ceiling. 

If the strips are to run lengthwise, find the number of strips 
by dividing the width of the room by the width of the paper, 
then proceed as in case of walls. 

To estimate for the border. 

Find the total perimeter of the room. Estimate cost per 
running yard. 

When double rolls are available they would be used, if 
more economical in cutting. 






APPLICATION OF MEASURES TO THE TRADES 69 

Example:— A room 16' 0" wide, 20' 0" long and 9' 0" high 
from base-board, has two doors each 4' 0" wide and three 
windows each 3' 6" wide. Find the cost of paper for the walls 
and ceiling, the wall paper being 18" wide and costing $2 a 
double roll, the ceiling paper being 18" wide and costing 80c. 
a double roll. 

Perimeter of room = (16' + 20') 2 = 72'. 

Perimeter — Width of doors and windows = 72' — 18£' = 53?'. 

Number of strips required = — ^^ — = 35f .".36. 

Number of strips in a double roll= -^- = 5^ .'. 5. 

Numberof rolls *£ = 7£.\8. Cost = $16. 

Number of strips required for the ceiling if running length- 

16X12 in2 . 
wise = — r^ — = lOf . . 11. 

lo 

Number of strips in a double roll = f£ = 2f .'.2. 

Number of rolls = V = o§ .'.6. Cost = $4.80. 

Total Cost $20.80. 

Painting. The area is usually estimated in sq. yd. 

The following is a common method of reckoning the area 
of doors, windows, etc.: 

Doors are taken to average 3' 0" X V 0", windows 3' 0" 
X 6' 0". If the window be divided into 12 lights the area is 
doubled, if divided into 6 lights one-half the area is added, 
and so on. The base-board is taken as 1' 0" by total perimeter, 
picture moulding 3" by total perimeter, and dado rail 6" by 
total perimeter. 

Exercises XXXIX. 

1. A room is 13' 0" wide, 15' 0" long and 8' 6" high. There 
are two doors each 2' 8" wide, and three windows each 3' 0" 
wide. Two of the windows have 6 lights and the other 2 
lights. A picture moulding 2" wide and a base-board 8" 
wide run around the room. Find (1) the cost of tinting the 
ceiling and 1' down to picture moulding at 25c. a sq.yd., (2) 
the cost of painting the interior woodwork at 50c. a sq. yd., 



70 MATHEMATICS FOR TECHNICAL SCHOOLS 

(3) the cost of papering the walls with paper 21" wide at 
$1.25 a roll, the decorator charging 40c. a roll for the work. 

2. A room 10' 8* wide, 11' 4" long and 8' 6" high, has two 
doors each 2' 10" wide, one window 4' wide, 2 lights, two 
windows each 3' wide, 12 lights, base-board 10" wide running 
around the room. Find (1) the cost of painting the woodwork 
at 25c. a sq. yd., (2) the cost of papering the ceiling with 
paper 18" wide at 25c. a single roll, (3) the cost of papering 
the walls with paper 30" wide at 90c. a roll, using a border 
4" wide at 20c. a yard. The decorator charges 30c. a roll for 
the work in both walls and ceiling. 

3. A room 12' 0"'wide, 18' 6" long and 10' 0" high, has two 
doors each 3' 10" wide, two windows each 2' wide, 4 lights in 
each, one window 4' 6" wide, 12 lights, a fire-place 5' 6" wide, 
a picture moulding 3" wide and a base-board 1' 0" wide running 
around the room. Find (1) the cost of tinting the ceiling and 
16" on wall to picture moulding at 30c. a sq. yd., (2) the cost 
of painting the woodwork at 40c. a sq. yd., (3) the cost of 
papering the walls with paper 18" wide at $1.20 a double roll, 
the decorator charging 50c. a roll for the work. 



CHAPTER VI. 
ALGEBRAIC NOTATION. 

63. In Arithmetic we denote quantities by numbers, each 
number having a fixed value. By 5 in. we mean that the line, 
or pencil, or bolt, is 5 in. in length. For this purpose we have 
the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. These symbols, in 
whatever way they are combined, have definite fixed values. 
In Algebra there is no limit to the number of symbols employed, 
the letters from our own alphabet being the ones chiefly 
used. 

64. Algebra — generalized Arithmetic. These symbols: — a, 
b, c, d, etc., in contrast to the symbols of Arithmetic, have not 
fixed values but may be given any values required by the 
conditions under discussion. Thus in Arithmetic 2X4 is 
always 8, whereas 2Xa, or more briefly 2a, will have different 
values according to the numerical values assigned to the 
symbol a. When a = 4, 2a = 2X4 = 8, when a = 8, 2a = 2X8 
= 16, and so on. Here the 2a is called an Algebraic Expression, 
the 2 being called the Numerical Coefficient, denoting the 
number of times a is taken in the sum. When no numerical 
coefficient is placed in front of the symbol, 1 is understood, thus 
a means la. 

65. Arithmetic Laws Applicable. Since the symbols a, b, 
c....x, y, z stand for numerical quantities we may apply 
the ordinary Arithmetic laws infusing them. In Arithmetic 
2X6+3X6 = 5X6 = 30. So in Algebra 2a+3a = 5a, 6a-2a = 
4a. In Arithmetic 2X6 = 6X2, so in Algebra aXb = bXa. 
Also 2X4X6=4X6X2 = 6X2X4, so in Algebra aXbXc = 
aXcXb = bXaXc = abc = acb = bca. If we then wish to add 
3afec+2ac6+7ca6 we should rearrange the terms thus, 
3abc+2abc+7abc = 12abc. An important difference between 

n 



72 



MATHEMATICS FOR TECHNICAL SCHOOLS 



the notation of Arithmetic and that of Algebra should be 
noted. In Arithmetic 31 means thirty-four or 3X10+4; in 
Algebra ab means aXb. 



Exercises XL. 

Find the values of the following: 

1. 3x+9x. 

2. a-\-a. 

3. 2a — a. 

4. 7x-3x. 

5. llx— 4a\ 

6. x— x. 

7. 3a6+5a6. 



8. 2a6+36a. 

9. ab — ba. 

10. Wxy — lxy. 

11. 9xy — 3yx. 

12. 6a6 — 6a. 

13. 8a6c — Scab. 

14. 3x+4x+5x. 



What is the value of 8x when: 

21. x = 2. 

22. x = 4. 



23. x= i 



24. x=-4. 



What is the value of -^ when : 



27. z= 4. 

28. a; = 16. 



29. x = 5. 

30. x = \. 



What is the value of -^ when: 
o 



33. x= 6. 

34. x = 18. 



35. x = 7-5. 
36.' a: = 2-7. 



15. 3a6+2a6+4a6. 

16. 5xy +Qxy+3xy. 

17. 3a6c+26ca+10ca6. 

18. x-\-x+x+x. 

19. 3x+4x+x+6x. 

20. 96+36+56+66. 



25. x= I 

26. x = 2\. 



31. x= -5. 

32. x = 2-b. 



37. ar=-6. 

38. x=-9. 



39. x= -036. 

40. x=-0024. 



Exercises XLI. 

1. What is the number which is 2 greater than x? 

2. What is the number which is 3 less than x? 

3. If an article costs x cents what is the cost of three articles? 
of seven articles? of eleven articles? 

4. Express x sq. ft. in sq. in. 

5. Express x sq. in. in sq. ft. 

6. Express x metres in (1) decimetres, (2) in centimetres, 
(3) in millimetres, (4) in kilometres. 

7. Express x millimetres (1) in centimetres, (2) in decimetres, 
(3) in metres, (4) in kilometres. 



ALGEBRAIC NOTATION 



73 



8. If there is an average of x trains leaving Toronto every 
day and an average of y cars per train, how many cars leave 
Toronto per day? 

9. In a rectangle ABCD if AB is c ft. in length and BC, b ft. 
in length, find (1) the perimeter of the rectangle, (2) the area 
of the rectangle. 

10. If the side of a square is b feet, find its perimeter. 

11. A can do a piece of work in m days and B in n days; 
write down (1) the amount of work each can do in 1 day, (2) 
the amount of work both can do in 1 day. 

12. The sides of a triangle measure x, y, z ft. Write 
down an expression for (1) the perimeter, (2) the semi- 
perimeter. 

13. A merchant mixes x ib. tea worth z c. a lb. with 
n lb. worth yea lb. Find the value of one lb. of the 
mixture. 

14. If a man works x hr. per day and handles y castings 
per hour, how many castings does he handle each day? 

15. If there are x cars in a railroad yard, how many 
trains will there be if there is to be an average of b cars 
per train? 

16. What is the length of the casting in the accompanying 
figure? 



Fig. 21 



17. What is the length of the casting in the accompanying 
figure? 



<f— 



Fig. 22 



74 



MATHEMATICS FOR TECHNICAL SCHOOLS 



18. Find the length of the crank-pin in the accompanying 
figure. 





< 




s« 




«. 














r 


- W. J 




-w » 


r 




■ 


* 




J 





Fig. 23 



19. If I is the length of the crank-pin in the accompanying 
figure what is the length of the last step? 













I 
























c 






s 








^ 






f 




r 


<' 












J 











Fig. 24 



20. If I is the length of the cylinder and saddle shoulder 
bolt in the accompanying figure what is the length of the 
shoulder? 




Fig. 25 

66. Index or Exponent, Power. In Arithmetic 3X3 may 
be written 3 2 or 9. 

In Algebra if we multiply a by a we cannot write the product 
as a single symbol, since we do not know the value of a; but 
we may express it as a 2 . In a similar way aXaXa^a 3 , 
aXaXaXa^a*. The small figure placed to the right and above 
the symbol is called the Index or Exponent, and the product a* 
is called the fourth power of a or more commonly a to the 
fourth. 



ALGEBRAIC NOTATION 75 

67. Index Laws. x i Xx 5 = xxxxxxxXxxxxxxxxx = x 9 . 
From this example we have the law: — The index of the 

product of two powers of the same symbol is equal to the sum 
of the indices of the factors. 

Examples : — a 6 X a 5 = a £+ 5 = a 11 . x 3 Xx 7 = x 10 . 
a 2 bXb 2 a = a 2+1 Xb 1 + 2 =a 3 b 3 . 

, . a 5 aXaXaXaXa 

a 5 + a 3 = —. = — — = a 2 . 

a 6 aXaXa 

If we cancel 3 of the factors in the numerator by the 3 

factors of the denominator, the above expression becomes 

aXa = a 2 . From this example we infer the law: — The index 

of the quotient of one power of a symbol divided by another power 

of the symbol is obtained by subtracting the index of the divisor 

from the index of the dividend. Examples: 

a 7 + a* p a 1 ~ i = a 3 . 6 25 -r-6 14 = 6 25 - 14 = 6 u . 

68. Some expressions in detail, a 3 means that a is taken 
three times as a product. 

ab 2 means that a is taken once, b is taken twice as a product, 
and the two results are multiplied together. 

6a 2 6 3 means that a is taken twice as a product, b is taken 
three times as a product, the two results are multiplied together, 
and the resultant product is taken six times. 

Exercises XLII. 
What is the: 

1. second power of a? 6. product of a 2 and a 3 ? 

2. third power of 4? 7. product of 4a and 36? 

3. fifth power of 5? 8. product of 4a 2 and 5a 3 ? 

4. sixth power of 6? 9. product of 12a6c and 3a6c? 

5. product of x and x 2 1 10. product of 3a 2 6 and a& 2 ? 

11. Express the product abx 2 in different forms. 

12. Do the same with Sx 2 y 3 , 6a 2 6 3 c 4 , 12ab 3 x. 

13. Write in detail what is meant by the following: 

4a 3 6, 5a 2 6 2 , 6a6c, 7a 3 6c. 



76 MATHEMATICS FOR TECHNICAL SCHOOLS 

Find the results of the following expressions in the most 
simplified form : 

14. a+3a-f-6a. 16. 3a6c+6a6c. 

15. 7a-2a+4a. 17. 7a6 + 106a. 

18. 13a6c+66ca — 2cab. 

19. 4:xyz-{-Qyxz-t-2zxy — £zyx. 
What is the result of: 



20. 


a 6 4- a 2 . 


24. 


xy-r-x^y 4 . 28. 


a 2 b 2 + abXa 3 b 3 . 


21. 


x 16 -i-x 3 . 


25. 


3a 2 6 2 H-a6. 29. 


Sx 3 y 3 -7-xyX 2x 2 y 2 . 


22. 


x 3 y 3 -±xy. 


26. 


15x 3 i/ 3 -7-a;V. 30. 


4:m 3 n 3 X rnn + wi 2 n 2 . 


23. 


x 2 +x 6 . 


27. 


x 2 y 2 Xx 3 y 3 s-xy. 





31. The side of a square is 6 in. What is its area? 

32. The edge of a cube is b in. What is the area of a face? 
What is the area of all the faces? What is the volume of the 
cube? 

33. The volume of a cube is 8x 3 . What is the area of a 
face? What is the area of all the faces? 

34. If a train travels I hr. at k miles per hr. and c hr. at d 
miles per hr., find the total distance travelled. 

35. Represent three consecutive numbers, (1) if x is the 
first one, (2) if x is the middle one, (3) if x is the last one. 

36. If the length of a stick is b ft. find its length in in., 
in yd., in rods. 

37. If a rod is x yd. b ft. and c in., how many inches in length 
is it? 

38. If x is the price per quart for beans, what is the price 
per gallon? What is the price per bushel? 

39. A man earned $x per day and his son $y. How many 
dollars did they both earn in a month if the man worked 25 
days and the son 20 days? 

69. Roots. As in Arithmetic the square root of x, or the 
expression whose second power is x, is indicated by y/x. 
Similarly the cube, fourth, fifth, etc., roots of x, or the expres- 
sions whose third, fourth, fifth, etc., power is x, are indicated 

by -s/x, \Zx, \Zx, etc. 









ALGEBRAIC NOTATION 



77 



Thus, ^a 6 =a 2 Since a 2 X a 2 X a 2 = a 6 . 

^a 12 = a 3 Since a 3 X a 3 X a 3 X a 3 = a 12 . 
^32=2 Since 2X2X2X2X2 = 32. 

The symbol V is called the radical sign. 
Exercises XLIII. 



Find the square root of: 

1. x 2 . 6. 16a 6 . 

2. x 6 . 7. 49x 2 # 2 . 

3. 16x 2 . 8. 81a 4 6 4 . 

4. x 12 . 9. 144a 6 6 6 

5. 64x 4 . 10. 169aV 

Find the value of: 

16. y/a 4 b\ 

17. V« 6 . 

18. V49-V36. 

19. V49 + V4. 

20. Vx 6 -x 3 . 

21. y/x*-x\ 

22. ^x s . 

23. ^/x 16 . 



11. 



12. 



13. ~ 



a? 

4* 

a 6 

9' 

x 8 

f6' 



14. 



15. 



a 2 b 2 

9 " 
x 4 y 4 z 4 

16 ' 



25. Square of 4x?/. 

26. Cube of x 2 . 

27. Fourth power of y 2 

28. Cube of 2 a 2 ?/ 4 . 

29. Cube root of x 6 . 

30. Cube root of 8a 3 . 

31. C ube root of 27a 6 . 

32. V25-16. 



34. V25a 4 -16a 4 . 

35. Vfi 

36 -€ 



24. Square of a 4 6. 33. V49-33. 



\a 16 . 



38 



70. Like and Unlike Terms. Two terms which contain the 
same letters involved in the same way are called like terms. 
Thus 6a and 3a are like terms. 3a6 and 4a6 are like terms. 
7x 2 and 9x 2 are like terms. 

Since ab and ba both mean aXb, ab and ba are also like 
terms, also 5a6 and 76a are like terms. 

Like terms may then be defined as terms that differ only in 
their numerical coefficients. 

Unlike terms may be defined as terms that differ in other than 
their numerical coefficients. 

Thus 6a and 46 are unlike terms. x 2 y and xy 2 are unlike 
terms. 7a 2 and 96 2 are unlike terms. 



78 MATHEMATICS FOR TECHNICAL SCHOOLS 

If we wish to add such terms all we can do is to write them 
down with a plus sign between them, thus 6a +46, x 2 y-\-xy 2 , 
7a 2 +96 2 . 

When we wish to simplify an algebraic expression such as 
3a+46 — 2a +66 we can combine the like terms 3a and — 2a, 
giving a, and the like terms 46 and 66, giving 106, and write 
the result a+106. 

Examples : 

10a+66-3a+4c-26-c = 10a-3a+66-26+4c-c 

= 7a+46+3c. 
9xy + 4x 2 ?/ 2 + 2xy — 3x 2 y 2 = 9xy + 2xy + 4x 2 2/ 2 — Zx 2 y 2 

= llxy-\-x 2 y 2 . 

Exercises XLIV. 

Simplify by combining like terms: 

1. 4a+3a+6a-2a. 5. 7xy+6x 2 y 2 -3yx+4y 2 x 2 +3xy- 2x 2 y 2 . 

2. 3a+2a + 66 — 46. 6. 3ra + 2rc+2m — m — n-\-3mn — n-\-2mn. 

3. 3a6+46a+36c-6c. 7. 6p+2g+4r-3p+6g-2r+4p-2g. 

4. 6a6c+3a 2 6 2 -26ca. 8. 3a+2x-4?/+7a+82/+5x. 

If a = 8, c = 0, k = 9, x = 4, y = l, find the value of: 

VcyK 12. 2x^2ay. 
25a 



9. 


V2ak 2 . 


10. 


yzk. 


11. 


13. 


5y\/4:kx. 


14. 


Sc\/kx. 


15. 


17. 




18. 


Ikax 2 

\i8y. 





v^ -v 2 



71. Brackets. In Arithmetic when a number of terms is 
included within a bracket it is understood that these terms 
are to be regarded as a whole. 

Thus, 10+ (5 +4) means that we first add 5 and 4 and then 
add the result to 10. Also 10 — (5+4) means that we first 
add 5 and 4 and then subtract the result from 10. So in 
Algebra, a + (6+c) means that we first add 6 and c and 
then add the result to a. 






ALGEBRAIC NOTATION 79 

Certain rules are necessary with respect to the signs of the 
terms within the bracket when the bracket is removed. These 
rules may be obtained by an analysis of a few type cases. 

By a-\-(b-{-c) we mean that the quantity b-\-c is to be added 
to a. We may first add 6 and then afterwards add c, giving 
a-\-b+c. By a-\-(b — c) we mean that the quantity obtained 
by subtracting c from b is to be added to a. It is evident 
that if we add b to a, obtaining a-\-b, our result will be too 
great by c; we must therefore subtract c from a-\-b, obtaining 
a-\-b—c as a result. From these illustrations we infer the 
rule: — When a group of terms is contained within a bracket 
preceded by the sign + the bracket may be removed without 
changing the signs of the terms within. 

In a — (b-\-c) we have to subtract the sum of b and c from a. 
If we subtract b from a, giving a — b, it is evident that the 
result is too great and that it is too great by c; therefore we 
must subtract c from a — b, giving a — b — c. In a — {b—c) we 
have to subtract the result b — c from a. If we subtract b 
from a, giving a — b, it is evident that we have taken away too 
much, for we were required to take away only b — c. The 
result a — b is therefore too small by c, and we must add c to 
a — b, giving a — b-\-c. From these illustrations we infer the 
rule: — When a group of terms is contained within a bracket 
preceded by the sign — the bracket may be removed by changing 
the signs of the terms within. 

3x means x-\-x-{-x, similarly 3(a-f-6) means (a+6) + (a+6) 
+ (a+6) =3a+36. This would lead us to the rule: — 
The product of an expression, consisting of two or more 
terms and a single factor, is the sum of the products of each 
term of the expression multiplied by the single factor. 

Examples: 1. 3x — (a-\-b) =3x —a —b. 

2. la +(b+c)=7a +b +c. 

3. 9x 2 — (x— y) =9x 2 — x -\-y. 

4. Q(a+b+c) =6a +6b+Qc. 



80 MATHEMATICS FOR TECHNICAL SCHOOLS 

It is necessary to note the difference between 3a 2 and 
(3a) 2 . In 3a 2 we have to multiply a by a and take the result 
three times. In (3a) 2 we have to square the whole quantity 
3a, giving 3aX3a or 9a 2 . 

Examples: 1. (7a6) 2 = 7a6X7a6=49a 2 6 2 , 

2. (2a 3 ) 4 =2a 3 X2a 3 X2a 3 X2a 3 = 16a 12 . 

It is sometimes necessary to enclose with brackets part of an 
expression already enclosed within brackets. In such cases 
the pairs of brackets are made of different shapes — ( ), 
{},[]. Thus a-{6 + (c-d)}. 

The same rules with respect to the removal of brackets 
apply, it being usually best to begin with the inside pair and 
remove one pair at a time. In the example given we would 
first simplify thus, a— {6-f-c— d}. We would then remove the 
remaining pair and write the expression a — b — c+d. 



Simplify: 




Exercises XLV. 








- 


1. 3a + (4a- 

2. 15x-(6a 


-2a). 
:+3x). 




3. 
4. 


36- 
6a- 


-(2c 
-(4c 


t+4a). 
t + 2a). 



Prove the following by removal of brackets: 

5. 6-r-0r-2)-(3+4x) + (6x + l)=3.r-f2. 

6. (3x-2)-(4x+5) + Or+7)=0. 

7. (9a -b) + (3a -26) -(6a -56)= 6a +26. 

8. (x-r-6a)-(2a:-3a)-(a-6x)=5x-(-8a. 

9. 2(x + l)+3(l+x)+2(2+3x)=9 + llx. 

10. 3(2-a)+6(2a+7) + (a-42) = 10a+6. 

11. 2(a+6)-(2a-6)=36. 

12. 3(a+b-r-c)-(6+a-c)-(2c-2a-6)=4a+3&+2c. 

13. 2(3x+12)-f-3(x-f-4)-(8a;-12)=x+48. 
Simplify: 

14. 3{x-(2x-6x)}. 17. 3x 2 +x(x+S)+x 2 . 

15. {3a + (6a-2a)+4a}. 18. a- {6 + (c-6?)}. 

16. x+{2z-f-30r+2x)}. 19. a-[6- {a-(6-a)-f-6} -a] 



ALGEBRAIC NOTATION 81 

20. Enclose a — b+c — d — e+i in alphabetical order in 
brackets, two letters in each; three letters in each. 

72. Negative Quantities. We have in Arithmetic found 
the value of an expression such as 6 — 5. In every case however 
the number to be subtracted was less than the number from 
which it was subtracted. , 

A difficulty is presented if we are asked to find the value of 
5 — 6. This is arithmetically impossible. We cannot take 
$6 from $5. We may, however, by making use of brackets, 
write 5 — 6 thus, 5 — (5 + 1) =5 — 5 — 1 . Here 5 — 5 is 0, and the 
value appears as 1 to be subtracted with nothing from which 
to subtract it. We shall say that the result is the negative 
number 1 or minus 1, and denote it by — 1. The idea of negative 
numbers may be made clearer by means of a graphical repre- 
sentation. 

I + + + + + + 



654321 123456 

In the above diagram we have represented numbers to the 
right of the vertical line as positive, and numbers to the left 
of the vertical line as negative. The two series of numbers 
may be considered as forming but a single series consisting of a 
positive branch, a negative branch, and zero. 

If then we wish to subtract 4 from 2 we begin at 2 in the 
positive series, count 4 units in the negative direction (to the 
left) and arrive at — 2 in the- negative series, that is, 2 — 4 = -2. 

A few examples may be added to show the practical value 
of negative quantities. 

Example 1 : — If the temperature is 30° below zero it may be 
recorded -30°. If it rises 5° it is then 25° below zero or -25°. 
If it increases 10° more it is 15° below zero or — 15°. 

Example 2: — If a merchant during a day's transactions gains 
$80 on one class of goods and loses $100 on another class we 
can represent the result of the day's business as $80 — $100 = 
-$20. 



82 MATHEMATICS FOR TECHNICAL SCHOOLS 

Example 3 : — If a man rowed 50 yards up stream and then 
drifted down 60 yards, his position relative to the starting 
point would be 50 yards — 60 yards = — 10 yards. 

Exercises XL VI. 

1. A man has $500 and owes $500. How much is he worth? 

2. A man has $500 and owes $700. How much is he worth? 

3. A man goes 5 miles north of Barrie, then 9 miles south. 
How many miles north of Barrie is he? How many miles 
has he travelled? Make a diagram showing his route and his 
last position. 

4. The temperature at 6.00 A.M. is +14° and during the 
morning it grows colder at the rate of 4° an hour. Find the 
temperature at 9.00 A.M., at 10.00 A.M., and at noon. 

5. A freight engine is switching in front of a station. If 
it runs 400 ft. to the right of the station (+400 ft.) and then 
backs 525 ft. (— 525 ft.), how many feet is it from the 
station? 

6. In drilling a well the drill is raised 8 ft. (+8 ft.) above the 
surface. It is then dropped 15 ft. ( — 15 ft.). Where is it 
then with respect to the surface? 

7. A boy is fishing in deep water with a line 20 ft. long. If 
the tip of the pole is +6 ft. above the water, how far is the 
sinker from the surface of the water, if it is 3 ft. from the hook? 

8. A man who was $350 in debt contracted another debt 
of $200. He then earned $1000. How much was he then 
worth? 

9. A boat, that runs 16 miles an hour in still water, is going 
against a stream flowing 4 miles an hour. What is the rate 
at which the boat travels? 

10. If a mine is opened 200 ft. above the base of a mountain 
and a shaft is sunk 700 ft., how much is the base of the shaft 
above or below the base of the mountain? 

11. A man starts from a point on a road running north 
and south, and walks c miles north and then 6 miles in the 
opposite direction. How far is he now from the starting 
point? How far has he travelled? 

Illustrate from the following cases: — (1) c = 8, 6 = 6. 
(2) c = 5, 6 = 9. (3) c = 8, 6 = 8. 



ALGEBRAIC NOTATION 83 

12. The thermometer stands at x°; in the course of an 
hour there is a fall of y° and in the course of the next hour a 
rise of z°. Find the reading at the end of this time. 

Illustrate for the following cases: — (1) x = 6, y — 4, z = 5; 
(2) x = -8, = 4, z = 9; (3) x= -2, = 3, z = 6. 

Simplify: 

13. 3a+2&-4a+66-8a-9&. 

14. 2s-3s+s-s-5s+5s. 

15. 4.r 3 -ox 3 +3x 3 -8x 3 +7x 3 . 

16. -56 +§6-f& +26 -!&+£&. 

17. fx-|0+|a;+f0-|x+|0-0. 

18. 6a 2 -3a 2 +2& 2 +3a 2 -2& 2 . 

19. x 2 +xy+y 2 -3x 2 -2xy+4y 2 . 

20. 3p+25-2p+4g-6p. 



CHAPTER VII. 



SIMPLE EQUATIONS. 

73. We might say that the greater part of a student's work 
in Arithmetic has been concerned with equations. The state- 
ment that 3 added to 4 is 7 might be expressed in the form 
3-|-4 = 7. This is an equation or a statement of equality 
between two expressions, 3+4 being one and 7 the other. 

All such equations involving only simple Arithmetical 
operations may be called Arithmetical equations, to distinguish 
them from equations of the form Sx = 9 which we will call 
Algebraic equations. 

The x in this equation is called the unknown and the process 
of finding its value is called solving the equation. 

An equation, in which the unknown quantity is involved to 
the first power only, is called a simple equation. 

In the given case if 3x = 9, then x = 3; the value 3 is said to 
satisfy the equation. 




D" 



O 



Fie. as 



74. Operations on the equation. The two sides of an 
equation must always balance, just as the weights in the 
two pans of the scales above must be equal if the scales are to 

84 



SIMPLE EQUATIONS 85 

balance. In the equation 3x = 9 if we add 2 to the left-hand 
side we must of necessity add 2 to the right-hand side. The 
equation then becomes 3x +2 = 9 +2 or 32+2 = 11. 

In the same way, if we subtract 2 from the left-hand side 
we must subtract the same quantity from the right-hand 
side. The equation then becomes 3a: — 2 = 9 — 2 or 3x — 2 = 7. 

Further, if we multiply the left-hand side by 2 we must 
multiply the right-hand side by 2, giving 2X3z = 2X9 or 
Qx = 18. 

We might also divide the left-hand side by 2 giving fx, 
but we would also have to divide the right-hand side by 2, 
giving f or the equation f#=f. 

75. Transpositions. Let us consider the equation 3x+4 = 16. 
By the previous paragraph we could subtract 4 from both 
sides of the equation, giving 3x+4 — 4 = 16— 4 or 3x = 16— 4. 
We then observe that the equation 3x+4 = 16 is equivalent 
to 3x = 16 — 4, or that the 4 has been moved from the left-hand 
side to the right-hand and that its sign has been changed. 

Next let us consider the equation Sx — 2 = 10. We could add 
2 to both sides giving 3x-2+2 = 10+2 or 3x = 10+2. We 
then observe that the equation 3^ — 2 = 10 is equivalent to 
3z = 104-2, or that the 2 has been moved from the left-hand 
side to the right and that its sign has been changed. These 
two examples would lead us to make the statement: — A quantity 
may be transferred from one side of an equation to the other 
without altering the balance, provided we change the sign of the 
quantity transferred. 

Examples: 

1. Solve the equation 3x-2+5x-4 = 3:r-10-7x + 16. 
Transposing so that we have all the terms containing x on 
the left and the other terms on the right we get 

3x+5x-3x+7x= -10 + 16+2+4 giving 15x-3x = 22-10. 
or, 12x = 12. 
x = l. 



86 MATHEMATICS FOR TECHNICAL SCHOOLS 

2. Solve the equation 3(3z + l) -(x-1) = 6(x + 10). 
Multiplying out 9x+3-x + l =6x + 60. 

Transposing 9x— x — Qx = 60 — 3 — 1. 

2x = 56. 
x = 28. 
Verification: — Substitute the value 28 for x in the equation 
and we get 3(84 + 1) -(28-1) =6(28 + 10) 
or, 3X85-27 = 6X38 
255-27 = 6X38 
228=228. 

76. Need of the equation. Let us work the following 
problem, first without employing Algebraic symbols and 
then by making use of the Algebraic symbols, and compare 
the methods. 

Example : 

A shopper bought three articles, the second costing three 
times as much as the first and the third $3 more than the 
second; find the cost of each if the total cost was $10. 

First solution: — Suppose that the third article had cost as 
much as the second, then the total cost would have been 
$10— $3 or $7. Then for every share allotted to the first article 
we must allot three to the second and three to the third. This 
makes seven shares into which we must divide $7, giving $1 
for one share. 

/. the first article cost $1, 
the second article cost $3, 
the third article cost $3 +$3 = $6. 



Second solution: — Let x = No. of dollars in cost of first, 

then 3z = No. of dollars in cost of second, 
and 3x+3 = No. of dollars in cost of third, 
thena;+3a;+3a:+3 = 10 
7z+3 = 10 
7x = 10-3 
7x = 7 
x = l 
3x = 3 
3z+3 = 3+3 = 6. 



SIMPLE EQUATIONS 87 

or the first article cost $1, the second $3, the third $6. If we 
compare the two solutions it is evident that the latter method 
has the advantage in both directness and clearness. 
Additional examples: 

1. A man works a full day of 8 hours, and in addition works 
3 hours overtime, for which he receives time and a half. If he 
is paid $8.75 for the entire time, what is his regular rate per 
hour? 

Let x c. = regular rate per hour, 

then f x c. =rate per hour for overtime, 
then 8 x c. =pay for 8 hours' work, 
and 3 Xf xorfx c. =pay for overtime, 
.'. &r+fa: = 875. 
Multiplying both sides of the equation by 2 we get 
16x+9x = 1750 
25x = 1750 
a; = 70, 
or the regular rate is 70c. per hour and the overtime rate is 
| X 70 or $1.05 per hour. 

2. How much water must be added to a quart of alcohol, 
which already contains 5% of water, so that the mixture may 
contain 50% of alcohol? (No allowance being made for con- 
traction). 

Let x =the number of quarts of water to be added, 

then 1+x = total number of quarts of mixture, 
and § (l+x) = number of quarts of alcohol in the mixture. 
Since no alcohol has been added, this must equal the number 
of quarts of alcohol in the mixture at the beginning. 

Multiplying both sides through by 100 we get: 
50 (l+x) =95 
50+50* =95 
50x =45 

/- _ 45 _ 9 

x — S1J — TO - 

.'. ^ quarts of water must be added. 



88 MATHEMATICS FOR TECHNICAL SCHOOLS 

3. The sum of $1100 is invested, part at 5% and part at 6% 
per annum. If the total income is $59, how much was invested 
at each rate? 

Let $x = amount invested at 5% 

then income = T % li $x 

and $(1100— x) = amount invested at 6% 

then income = T |f 7y $(1100— x) 

ordfos+dhr (1100-x)=59. 

Multiplying through by 100 we get: 

5x +6600 -6x = 5900 

6600- 5900 = 6x-5x 

700 = a: 

.*. $700 invested at 5% and $400 at 6%. 

Exercises XLVII. 

Solve the following and verify : 

1. 3x+x = 64. 17. -2x =4. 

2. 5x+4x = 81. x X 

3. 8x-3x = 50. 3~2* 

4. 13x-3x = 100. Z^_oi 

5. 4x+7 = 3x + 10. iy - 9 " ■ 

6. 9x-6 = 7x-4. 2Q 3 = _x_ 

7. 7x+3=3x+67. * 4 12* 

8. 2x-7 = ll-4x. 21 H^_19f = 
0. 27-3x = 68-4x. ' 13 31 

10. 42-3x=48-9x. 22 *-3 

11. 6x-18 = 4x-8-3x+5. ' 5 

12. 10x-10-6x-27 = 3. „, 2x-l 

13. 24x + 10-20x + 100 = ' 3 

5x+96. 3x+5 = 

14. 5x=-05. ' 7 

15. 8x=-24. 25. f (x-10)=0. 

16. -7x = 21. 26. i (6x-15)=0. 



SIMPLE EQUATIONS 89 

27. 3 (3z + l)-(z-l)=6 O + 10). 

28. 3 (2a: + 5)-(4x-12)=5 (3x + l)-4. 

29. (lLr-22)-(8-6x)-(4-8x)=17a:+7. 

30. z(x+4)=x 2 +36. 

31. x 2 +2:r=z 2 +4. 

32. 3z 2 -5-(3x 2 -a:)=0. 

33. f +|-i-;+|. 

4 5 4 o 2 

«u -g+3 y+5_g+9 x+4 13 

4 + 2 8 + 3 "^12* 

q* *+ 3 « ,x + l_a-+5 1 

__ x x + 2 3+x 

36 - 3 +_ T = ^- 

37. ^+3* *~+2*+9. 

38. -09x-01x=-14--06a:. 

39. -03:r+-02=-17. 

40. • 007a: -• 008 =-004x+- 412. 

' -25x+-025 2z+-45 . a 
42 ' , -125 = -T^ + - 6 - 

Exercises XL VIII. 

1. A tree 84 ft. high is broken so that the length of the part 
broken off is five times the length of the part standing. What 
is the length of each part? 

2. After selling \ of his farm and then \ of what was left a 
man still has 140 acres. How many acres had he at first? 

3. The length of a rectangular building is b ft., the width is 
70 ft., and its area is 43,400 sq. ft. Find the value of b. 

4. A rectangular building is 84 ft. long and x ft. wide. Find 
x if the area is 13,440 sq. ft. 

5. A rectangular shop is x ft. long and y ft. wide. If x = 120 
and y = 48 what is the area? 



90 MATHEMATICS FOR TECHNICAL SCHOOLS 

6. The desired area of a new rectangular boiler shop is 
x sq. ft. Owing to the space available the width is limited to 
b ft. What must be the length c, if 6 = 64 and x = 9648? 

7. The length of a rectangular machine shop is x ft., the 
width 50 ft., and the floor space must be capable of accommo- 
dating 20 machines, each occupying an average of 300 sq. ft. 
Find the value of x. 

8. The front section of an engine frame is required to 
have 60 sq. in. area, the width is 5 in., and the depth is x in. 
Find x. 

9. A man saves $100 more than | of his salary, spends 4 
times as much for living expenses as he saves, and pays the re- 
mainder which is $500 for rent. What is his salary? 

10. If air is a mixture of 4 parts of nitrogen to 1 part of 
oxygen, how many cubic feet of each are there in a room 20 
ft. by 30 ft. by 10 ft.? 

11. The length of a room is to its width as 4 is to 3 and its. 
perimeter is 70 ft. Find the width of the room. 

12. The number-plate on an automobile has a perimeter of 
48 in., and its length is to its width as 3 is to 1. Find its length 
and width. 

13. Sirloin steak costs 1| times as much as round steak. 
Find the cost per lb. of each if 3 lb. sirloin and 5 lb. round steak 
cost $3.04. 

14. If 2 lb. butter cost as much as 5 lb. lard, and 4| lb. lard 
and 6 lb. butter cost $5.07, find the cost of each per pound. 

15. The interest on $138 for a certain time at 6% per annum 
is $16.56. Find the time. 

16. A can do a piece of work in 6 days, B can do the same 
work in 8 days, and C in 24 days. In how many days can they 
do the work if they all work together? 

17. A tank is emptied by two pipes; one can empty the tank 
in 30 min., the other in 25 min. If the tank is f full and both 
pipes are opened, in what time will it be emptied? 

18. Four pipes discharge into a cistern; one fills it in one day, 
the second in two days, the third in three days, the fourth 
in four days. If all run together bow soon will they fill the 
cistern? 



SIMPLE EQUATIONS 91 

19. A train runs 100 miles in the same time as a second 
train runs 120 miles. If the rate of the first train is 5 miles 
an hour less than that of the second train, find the rate of 
each. 

20. How many quarts of water must be mixed with 250 
quarts of alcohol 80% pure to make a mixture 75% pure? (No 
allowance for contraction). 

21. A man sells \ his interest in a factory and later sells 
\ of what he has left. His interest is then worth $75,000.00. 
How much was his original interest worth? 

22. A lot of brass scrap weighing 500 lb. contains 25% zinc. 
How many pounds of zinc must be added in melting to increase 
the percentage of zinc to 34%? 

23. From a tank one-half full of crude oil, 500 gallons are 
drawn and 25 gallons are lost by evaporation and leakage. 
If the tank is then one-quarter full, how much does it hold 
when full? 

24. The sales of a firm increased 10% the second year 
over the first, and the third year they were 20% more than 
they were the second year. If the sales total $235,125.00 
the third year, how much were they for the first year? 

25. What is the value of the property of a person whose 
income is $645.00, when he has two-thirds of it .invested at 
4%, one-fourth at 3%, and the remainder at 2%? 

26. If a boy weighing 75 lb. sits 6 ft. from the fulcrum, 
where should a boy weighing 100 lb. sit to balance the beam? 

27. A weight of 200 grams is placed 25 centimetres from 
the fulcrum. How far from the fulcrum must a weight of 
one-half a kilogram be placed to balance the beam? 



CHAPTER VIII. 

FUNDAMENTAL OPERATIONS. 

77. Addition. In a previous section we dealt with the 
addition of simple expressions such as 6a and 3a, 4x 2 and 
3x 2 , etc. We now wish to deal with compound expressions 
such as 2a+56, 6a — 4x+36 2 , etc. If we wish to add a number 
of these compound expressions we must recall what was 
stated with respect to like and unlike terms. It was there 
pointed out that like terms may be added, as for example, 
6a and 2a, giving 8a. It was also stated that unlike terms 
could not be added in the above way but merely written 
with a plus sign between them. Thus 6a 2 plus 76 2 would be 
written 6a 2 +76 2 . If then we wish to add 3a — 56+c, 
2a +46 — c and 66+ 7a — 2c greater accuracy may be secured 
by arranging so that the like terms would be in the same 
vertical columns. 

Thus, Example 1: 3a — 56+c 

2a+46-c 

7a+66+2c 
Sum = 12a +56+ 2c' 

Example 2: Add 5ax — 7by-\-cz, ax-\-2by—cz, 

-Sax + 2by+3cz. 

Arrange as above giving : 

5ax — 7by+cz 

ax-\-2by — cz 

— 3aar+26i/+3cz 
3ax — 36y+3cz 

92 



FUNDAMENTAL OPERATIONS 93 

Exercises XLIX. 
Add: 

1. x 3 -3x 2 , Sx 2 -4x, 4z + l. 

2. 3(x-l), 4 0-1). 

3. x — 2y+3z, 2x-\-y — 3z, x — 2y-\-z. 

4. a+6, a — 6. 
_ a b a.b 
5 * 2 + 2' 2 + 2* 

6. a — c, b—c. 

7. z 2 -2:n/+2/ 2 , x 2 +2xy+y\ 

8. x+y — 2, 3x — 2y+4z. 

9. z-(?/+z), 2/-(x-z). 

10. 4(z-*/), o(x-y), Q(x-y). 

Find the values of the following sums when x = \, y = \, z = \, 
a = 3, 6 = 2, c = \. 

11. §a-f-£&— c, a — j6 — §c, oa — |6+2c. 

12. 5xy — 5x 2 y — 5xy, \xy-\-^x 2 y. 

13. fa-ffc+fc, fa-i&+|c. 

14. 12j/z-8x?/+ia+|6c. 

78. Subtraction. In its most elementary form subtraction 
has already been dealt with in connection with like terms. 

Thus, 6a — 2a = 4a. 

7a — 9a = —2a. 

Also the rules for the removal of brackets would deal with 
an expression such as 6a — (—3a). We could write this 
expression 6a — (0 — 3a) =6a — 0+3a = 6a+3a. 

also, — 7x — ( — ox) = — 7x — (0 — 5x) = — 7x — 0+5a; 
= — 7x + 5.r. 

An examination of the operation and the result in the two 
latter examples brings us to a very important result with 
respect to subtraction. In the first example we see that the 
subtracting of —3a from 6a is equivalent to adding -f-3a to 
6a; in the second that the subtracting of — ox from — 7.c is 
the same as adding +ox to — 7x. This gives us the funda- 
mental principle with respect to subtraction: — To subtract one 



94 MATHEMATICS FOR TECHNICAL SCHOOLS 

expression from another we change the sign of the quantity to be 
subtracted and add it to the other expression. 

An examination of the following examples in subtraction 
placed as in Arithmetic would illustrate this: 

6 4a 7x 2 Sab -6x 2 

9 2a 8x 2 -2ab -3s 2 

-3 +2a -x 2 +5a6 -3x 2 . 

If we wish to subtract one compound expression from 
another we arrange as in addition. Thus to subtract 3a — 26-f-c 
from 46 — 6a — 3c we write 

46 -6a -3c 
-26+3a+c 
66 -9a -4c. 

Exercises L. 
Subtract: 

1. 4a-36+c from 2a-36+c. 

2. a-36+5c from 3a-66+2c. 

3. 2x — Sy-\-z from 15?/ — 6x+4z. 

4. — 4:xy-{-2yz — lOzx from Zxy—6yz-\-7zy. 

5. 4x 2 -6x+2 from 7z 2 -3x-4. 

6. From the sum of 3a+26 and 7a — 36 subtract 3a — 6. 

7. Subtract 5x 2 +3x — 1 from 6x 3 and add the result to 
3z 2 +2x + l. 

8. Add the sum of 2y — Zy 2 and 1— 4y 3 to the remainder 
obtained when l—fy 2J t-2y is subtracted from 8?/ 3 +3. 

79. Multiplication. The method of representing the pro- 
duct of two simple expressions has already been given, thus 
the product of a and b = ab, the product of a, 6, and c = abc, 
the product of x, y, z, and k=xyzk. 

Combining this with our index laws we can find the product 
of expressions like x 2 y 2 and xy giving x 2 y 2 Xxy = x 3 y 3 . 
Also, 3x 2 X7x 2 = 3X7 Xx 2 Xx 2 = 21x* 

and, 4x 3 X- 2 = 4X2Xz 3 X-, = 8a; 3 - 2 = 8x. 



FUNDAMENTAL OPERATIONS 95 

In the section dealing with brackets it was seen that 
3(a+6) = 3a+36. In this case one of the expressions, 3, is a 
simple expression while the other a +6 is a compound ex- 
pression. 

If now we wish to multiply two compound expressions 
together, say x-\-a by x-\-b, we may write it in the form 
(x+a)(x+b). 

The work may be conveniently arranged thus, 
x + 6 
x-\-a 
x 2 -\-bx 

-\-ax-\-ab 
x 2 -\-bx-\-ax-{-ab. 

Multiply x +6 by x f then multiply x+b by a and add 
the results. 

Example: Multiply x+2 by x+3 
x+2 
x+3 
x 2 +2x 

+3s+6 
z 2 +5a:+6. 

80. Rule of Signs in Multiplication. In the examples 
given above all the signs are plus. It is necessary to consider 
cases where the signs are minus, or some plus and some minus. 

We might first recall the meaning of multiplication as 
understood in Arithmetic. The fundamental unit was + 1 
and all numbers were obtained from this unit. 

Thus, 3 = 1 + 1 + 1. 

Also, 3X4 = 3+3+3+3. 



96 MATHEMATICS FOR TECHNICAL SCHOOLS 

From this multiplication might have been denned as 
follows: — To multiply one number by a second is to do to the 
first what was done to unity to obtain the second. 

This law applies with equal force to the multiplication of 
fractions. Thus to multiply f by f we do to f what was done 
to unity to get f : that is, we divide f into four equal parts 
and take three of them. Each part would be ^j, and 
by taking three of these parts we get -£- X 3 = | X f. 

We will, therefore, make the above definition the basis of 
the rule of signs in multiplication. 

(1) To multiply+3 by+4, 

+3X+4 = +3+3+3+3 = +12, 

or generally +aX+&=+a&. 

*(2) To multiply -3 by +4. 

If we do to — 3 what was done to unity to obtain 4 we 
have -3 X +4 =-3 -3 -3 -3 =-12, 

or generally — a X +6= —ab. 

(3) To multiply+3 by -4. 

To obtain —4 from the fundamental unit we changed its 
sign and took it four times. If this be done with +3 then 

+3 X -4 =-3 -3 -3 -3 =-12, 

or generally +aX — b= — ab. 

(4) To multiply -3 by -4. 

Explaining —4 as in (3) and applying definition we have 

-3X -4= +3+3+3+3 = +12, 

or generally — a X — b= -\-ab. 

The results of (1), (2), (3), (4) may be stated in words 
giving the following rule for signs in multiplication: — The 
product of two numbers with like signs is positive and with 
unlike signs is negative. 



FUNDAMENTAL OPERATIONS 97 

Exercises LI. 

Multiply: 

1. 3a by 2. 7. 7x 3 by-3z. 13. x 2 y 2 z 2 by-xyz. 

2. Sx by -2. 8. a 2 b by -ab. 14. fcrby-fy. 

3. -26 by -4. 9. 4a: 2 by -2x. 15. fa 2 by -|6 3 . 

4. -3a 2 by a 2 . 10. p 3 by-p 2 . 16. §x 3 by-fx 2 . 

5. -3a6 by 2a6. 11. a 3 6 by-a6 3 . 17.±x 2 yby-&xy 2 . 

6. Zxbyly. 12. p 11 by -p 3 . 18. - 1 3 T a6 2 by^-a 2 6. 

19. \x 2 y 2 by ^. 20. -4:X 2 y by-5x 3 y. 

x y 

Write down the continued product of: 

21. -3,-4,6. 25. 2a, 36, -a. 29. x, -x,x, -x. 

22. a, -6, c. 26. 2x, -dx, -4z. 30. 3p 2 , 2pq, iqp. 

23. a 2 , -6 2 , c 2 . 27. a 2 x, x, 2/. 31. 2x,-Sx 2 ,-2x i ,-x 5 . 

24. -6 2 ,-c 2 , a. 28. -2x,-2x,-2x. 32. a 2 , 6 3 , 2c. 

Write down the values of: 

33. (-x) 3 . 39. (2xy) 3 . 45. (-a; 3 ) 5 . 

34. (-a) 4 . 40. (-1) 2 . 46. (-2a 2 6) 2 . 

35. (-2a) 3 . 41. (-1) 3 . 47. (-3x 2 y) 3 . 

36. (x 2 ) 3 . 42. (-1) 4 . 48. (-3x 2 2/) 4 . 

37. (-a) 6 . 43. (-1) 5 . • 49. \-7x 2 y 2 ) 2 . 

38. (-x 2 ) 3 . 44. (-X 2 ) 7 . 50. l-xyz) 3 . 

Exercises LII. 

Multiply: 

1. a+6— c by 4. 6. a 2 — a6+6 2 by — a. 

2. 2a-36+c by -2. 7. 3z 4 -2;r 3 -f-6 by-5z. 

3. x+^+2 by 2i. 8. -3a 2 -2a6+6 2 by-26 2 . 

4. 3x 2 +y 2 by-2z. 9. l-2x-\-x 2 by-2x. 

5. x 2 -\-2xy-\-y 2 by x. 10. x 2 —y 2 by—xy. 

Find the continued product of: 

11. a+6, a, 6. 14. a — 6, a, —6. 

12. a 2 -2a6+6 2 , a, 6. 15. z 4 -3x 3 +2x 2 -l,-3x,-2x. 

13. x 2 -5z+3, x 2 , x. 16. a 3 -a 2 6+a6 2 -6 3 , -a, -6. 



98 MATHEMATICS FOR TECHNICAL SCHOOLS 

When a = — 2, 6= —3, find the value of: 

17. a 2 -2. 22. 6 4 -81. 27. a 2 +6-6 2 . 

18. 2a 2 -a+2. 23. 6 2 -a 2 +2a. 28. a 4 -b\ 

19. a 2 -b\ 24. a 3 + 8. 29. a 5 -6 5 . 

20. a 2 -2ao+6 2 . 25. a 3 +b\ 30. a 3 -36. 

21. 2a 3 + 16. 26. 8a 2 -6 3 . 31. a 4 -l. 

Exercises LIII. 

Find the product of : 

1. z + a, x — b. 11. x 2 — a 2 ,xA-a. 

2. ay -6, q/-d. 12. x + 2y, 3x + l. 

3. 5+3x, 7-2*. 13. 7a-2b, a 2 -b 2 . 

4. x — by, 2x+?>y. 14. ax 2 — bx, ax-\-b. 

5. a-f-3ar, a — ox. 15. 6a — 26, a — 6. 

6. az + 1, 6x + l. 16. a+6, c— d. 

7. 4a 2 -36, 2a 2 -6. 17. x 2 +a, x 3 -b. 

8. a: 3 — 1, x + 1. 18. bx — ay,ax—cy. 

9. a 2 +66, a 2 -46. 19. ary»-l,«y+2. 
10. a+3x, a-bx. 20. x 5 -l, x 4 +l. 

81. There is a number of types of products in which the 
results can be written down by inspection if a few typical 
examples are examined. 

(1) (x-\-a)(x — a) =x -\-a 
x — a 



x 2 -\-ax 



— ax — a 2 
x 2 —a 2 . 

That is, the product of the sum and difference of two quantities 
is equal to the difference of their squares. 
Thus, (x+3)(s-3) = x 2 -9. 
(a+b)(a-b)=a 2 -b 2 . 
(xy + l)(xy — l) =x 2 y 2 — l. 
(2) (a+6)(a+6) or (a+b) 2 = a +6 

a +6 
a 2 +ab 

+ab + b* 
a 2 +2ab+b\ 






FUNDAMENTAL OPERATIONS 99 

That is (a+6) 2 = the square of a, plus the square of 6, 
plus twice the product of a and b. Any expression consisting 
of two terms is called a binomial, so that we may state as 
a general rule: — The square of a binomial is equal to the sum 
of the squares of the terms plus twice their product. 

Thus, (x+3) 2 = x 2 +9+6x. 
(x-4:) 2 = x 2 +lQ-8x. 
(xy — iy = x 2 y 2 -\-\— 2xy. 

Also, (a+6+c) 2 ={a+(6+c)} 2 = a 2 +(6+c) 2 +2a(6+c). 
= a 2 +6 2 +c 2 +26c+2a6+2ac. 
= a 2 +6 2 +c 2 +2a&+2ac + 26c. 

This method may be used for the square of an expression 
containing any number of terms so that the rule may be 
given thus : — The square of an expression consisting of any 
number of terms is equal to the sum of the squares of each of 
the terms plus twice the product of each term multiplied by each 
of the terms that follow it. 

(3) (x +2) +3)= x +2 
x+3 
x 2 +2x 

+3x+6 
z 2 +5x+6. 

Here we observe that the first term in the product x* is 
obtained by multiplying the first terms in each of the factors, 
the second term 5x is obtained by adding the 3 and the 2 
and multiplying by x, the third term is obtained by multiplying 
the 2 and the 3 together. 

Thus, (x+4)(a:+5)=x 2 +9^+20. 

(x-4)(;r-f3)=.r 2 -x-12. 
(x-6)(x-4)=x 2 -10x-r-24. 



100 MATHEMATICS FOR TECHNICAL SCHOOLS 

Exercises LIV. 
Write down the results of the following: 

1. (c+d)(c-d). 14. (2x+3y) 2 . 

2. (2s +3) (2s -3). 15. (xy + l)\ 

3. (x 2 -2a 2 )(x 2 +2a 2 ). 16. (x 2 -l) 2 . 

4. (s 2 +2)(x 2 -2). 17. (a+6-c) 2 . 

5. (a+36)(a-36). 18. (2a-6-c) 2 . 

6. (px+q)(px-q). 19. ( a +b+c-d)\ 

7. (x 2 -3y 2 )(x 2 +Sy*). 20. (2a-36+c) 2 . 

8. (2x-Sy)(2x+dy). 21. (z+3)(s+4). 

9. (a 2 -46)(a 2 +46). 22. (a+5)(a-2). 

10. (*+*/) Or -?/)(:r 2 +2/ 2 ). 23. (z+8)(z-5). 

11. (c+d)*. 24. (p+3g)(p-6g). 

12. (a-26) 2 . 25. (a&+4)(a&-5). 

13. (2x-^) 2 . 26. (*y-6)(*y+c). 

Use the rule for the square of a binomial to find the 
value of: 

27. 99 2 . 29. 105 2 . 31. (100-6) 2 . 

28. 102 2 . 30. 95 2 . 32. (99-5) 2 . 

82. Division. The Rule of Signs in Division may be 

readily deduced from the rule in Multiplication. 

Thus, (1) +xy=+xx+y .'. + X y++x=+yor^p!- = +y. 

(2) -xy= -xx+y .*.. -xy+-x=+y or —^=+y. 

— x 

~\~xy 

(3) +xy= —vx-y .*. +xy-¥ —x= — y or — -f-~ — y. 

(4) — z?/ = +sx-2/ .". —xy-?-+x=—y or —^-=-y. 

+x 

From these results we have the following rule of signs in 
division: — Terms with like signs when divided give plus (+). 
Terms with unlike signs when divided give minus (— ). 



FUNDAMENTAL OPERATIONS 101 

Examples: q-s= +3. T7o = ~' - ^' 

+6 --3 ^--3 

±2la^ = _ 7a -Zx 2 y 2 ^ 3 

— 3a +xy 

-35a 3 6 2 c _ ,, + 5x 7 . 

= 5a 2 o. — ^-i=— a: 5 . 



— labc — 5x 2 

Exercises LV. 
Divide: 

1. 3xby 3. 8. -6 4 by 6. 

2. -3x by 3. 9. 8a 2 by -4a 2 , 

3. -3xby-3. 10. -54a 2 6c by 6abc. 

4. -3xbyx. 11. 24a 2 6 2 c 2 by-406c. 

5. 6xy by 6x. 12. -21xV by-7x 3 ?/ 2 . 

6. a 2 by-a 2 . 13. -49a 3 6 3 by 7a 2 6 2 . 

7. 8a 2 by -4a. 14. -x 5 by+x 2 . 

Simplify: 

15. A^ 18. 24y2 * 2 , 21 121 * V 



5 —4?/ llx 3 i/ 



— 21x 3 ?/ 3 49pa 2 r -16a 3 6 3 

— 3xi/ ' ' —7pqr' — 8a 2 b 

1? -8x^ 2Q -32Z. 2 m 2 r^ ^ \abc 



■xy ' 4:1m ' \abc 2 ' 

Divide: 

24. 3x-6?/ by 3. 30. 6a -96 + 12c by -3. 

25. 3x-9 by -3. 31. x 3 +3x 2 -3x by x. 

26. 3x 2 -6x by -3x. 32. 15?/ 4 -5?/ 3 x 3 -30?/ 3 by by. 

27. -6 2 +a6by6. 33. -5m 3 n+20m 2 w 3 by -5mn. 

28. 4a 2 6-8a6 2 by -2a6. 34. a 2 bc-ab 2 c-\-abc 2 by -abc. 

29. -x 3 +x 2 by-x 2 . 35. - a 2 b 2 c 2 + abc 2 - cab 2 by abc. 



102 MATHEMATICS FOR TECHNICAL SCHOOLS 

83. To divide one compound expression by another the 
work may be arranged by following the method of long 
division in Arithmetic: 

Example. Divide x 2 +5x+G by x+2. 
s+2)x 2 +5x+6 /x+3. 

x 2 +2x (1) 

3x+6 (2) 

3s+6. (3) 

x 2 -i-x = x .'. x is the first term in the quotient, (x+2) multi- 
plied by x gives x 2 -\-2x and we obtain (1). Line (2) is obtained 
by subtracting x 2 + 2x from the expression and bringing 
down +6. 3x divided by x = 3, .*. 3 is the second term of the 
quotient. 

(x+2) multiplied by 3=3x-}-6 and we obtain line (3). 
This when subtracted leaves no remainder and the quotient 
is x+3. 

This method may be applied to an expression of any number 
of terms, if care is taken to arrange the divisor and dividend 
in descending or ascending powers of some common letter, 
and to keep the remainder in each case in the same order. 

Exercises LVI. 
Divide: 

1. x 2 +7x + 12byx+3. 9. 25-30a+9a 2 by 5-3a. 

2. a 2 +3a+2 by a+2. 10. 4x 4 -49 by 2x 2 -7. 

3. a 2 — 3a+2 by a — 1. 11. x 2 -\-ax-\-bx-\-ab by x-f-a. 

4. x 2 — 5x — 14 by x+2. 12. x 4 -\-x 2 y 2 -\-y 4 by x 2 — xy+y l . 

5. 15x 2 - 26x4-8 by 5x-2. 13. a 3 +6 3 by a+6. 

6. 6-13a+6a 2 by 2-3a. 14. x b +bx*y + l0x z y 2 + \0x 2 y* + 

7. 4+4x+x 2 by 2+x. 5xy A +y 5 by x 2 +2xy+y 2 . 

8. x 2 -\-2xy+y 2 by x+y. 15. a 3 +6 3 +c 3 — 3a6cby a + 6+c. 



CHAPTER IX. 

FORMULAS. 

84. One of the most Valuable Uses for algebraic symbols is 
to express a scientific law in a short form. When such a law 
is expressed in algebraic form it is called a formula. For 
example, the area of a rectangle is equal to the length multiplied 
by the breadth. If we let A represent the area, I the length, 
and b the breadth, we could briefly represent this relation by 
the equation A=lb. 

If Z = 15 in., 6 = 10 in., then A = 15 X 10 = 150 sq. in. 

Again if A =lb., then l = -i- This is called solving for I. 

If ,4=200 sq. in., 6 = 25 in., then Z=^=8in. 

Further if A— lb., 6 = 7- This is called solving for 6. 

If ^1=400 sq. in., Z = 40in., then 6=^00. = 10 in. 

If a scientific law be stated in detail it is important to be 
able to express it as a formula. 

Example. 

To find the number of revolutions of a driven pulley in a 
given time, multiply the diameter of the driving pulley by its 
number of revolutions in the given time, and divide by the 
diameter of the driven pulley. 

Using D and d for the diameters, and N and n for the number 
of revolutions respectively, express the above as a formula. 

Exercises LVII. 

1. The cutting or surface speed, that is the number of linear 
feet measured on the surface of the work that passes the edge 
of a cutting tool in a minute, is found by multiplying the 
circumference of the work being turned by its R.P.M. Express 
the rule as a formula. 

103 



104 MATHEMATICS FOR TECHNICAL SCHOOLS 

2. The current flowing along a conductor is given by the 

E 

formula 1= „, where / is the current in amperes, E the electro- 
it 

motive force in volts, and # the resistance in ohms. 
Solve for E and #. 

If £=110, # = 220, find/. 
If/=2|, £ = 220, find R. 

If/ =-5, R= 75, find E. 

3. The resistance of a wire in an electric circuit is given by 
R=K -j, where R is the resistance, L the length of the wire, 
A its area in circular mils, K the resistance of 1 mil foot in ohms. 

Solve for K, L and A. 

IfZ = 3ft., ,4 = 1000, # = 10-5, find R. 

If # = 220, K = 10-5, L = 1000 ft., find A. 

It R= 10, # = 10-5, ,4=250, find L. 

4. The work done by any force is given by W = FS, where W 
is the work done, F the force in pounds, S the distance in feet 
through which the force acts. 

Solve for F and S. 

If F = 525 lb., S = 51ft., find W. 

If FT = 150, F = 5oz., findS. 

If W = 500, S = 800 ft., find F. 

5. The total resistance of a series circuit is given by 
R = R l -\-R 2 -{-R 3 , where R is the total resistance and R 1} # 2 , # 3 
are the resistances of the separate parts of the circuit re- 
spectively. 

Solve for R lf R 2 and R 3 . 

tt R,= 2, #2 = 4 , #3 = 9, find R. 

If #=12, #]=2, # 2 = 7, find # 3 . 

6. The indicated horse-power of a single acting engine is 
given by: 

PLAN 

33000' 
where P is the mean effective pressure on the piston in pounds 
per sq. in., L the length of the stroke in feet, A the effective 
area of piston in sq. in., N the number of strokes per minute. 





FORMULAS 






10E 


Solve for P, 


L, A and N. • 








If P =80, 


Z = 2ft., ^4=30sq. 


in., 


N= 60, 


find 77. P. 


Ifff.P.= 4, 


Z = Uft., ^l=24sq. 


in., 


N= 50, 


find P. 


If ff.P. = 10, 


p = 50, ^l=30sq. 


in., 


iV=100, 


find Z. 


If #.P.=20, 


p = 60, Z = 2 ft., 




# = 100, 


find A. 


If fl.P. = 16, 


P = 60, i = 2 ft., 




A= 40, 


find iV. 



7. The formula D=QIT is used in electrolysis, where D is 
the weight of the deposit, Q the electro-chemical equivalent, 
T the time in seconds, 7 the current. 

Solve for Q, I and T. 

If Q = -001118, 7=40, T = 600, find D. 

If 7)= 2, Q= -000328, 7 = 250, find T. 

If7)=-5, 7=10, Z" = 153, find Q. 

8. The diameter of a rivet is given by d = 1 • 2^/t, where d 
is the diameter in inches and t the thickness of the plate in 
inches. 

lit =-75, find d. 
If d = -£g in., find t. 

9. The space through which a body falls from rest is given 
by s = h 9t 2 > where s is the space in ft., g the acceleration due 
to gravity, t the time in seconds. 

Solve for g and t. 

If * = 12, g =32-2, find *. 

If 5 = 3155.6, = 32-2, find t. 

W 

10. In a machine E= 5-p,, where 7? is the efficiency, W the 

weight, V the velocity ratio, P the horizontal force. 
Solve for W, P and F. 

IfJF = 112, P = 20, F= 12 -5, find E. 

If£=-55, P = 25, F=18-5, find W. 

If E = • 74, IF = 350, K = 23, find P. 

If £=-346, JF = 799-26, P = 20, find V. 

11. The allowable working pressure in a steam boiler is 

givenby: „ 2Tsk 

DF ' 

* 

where T is the thickness of the plate in inches, s the tensile 
strength of plate in pounds per sq. in., k the efficiency of the 



106 MATHEMATICS FOR TECHNICAL SCHOOLS 

joint, D the inside diameter of shell in in., F the factor of 
safety. 

Solve for T, s, k, D and F. 

If T= h 5 = 35000, fc=- 45, D = 30, F = 4, find B. 

IiB = 75, s = 40000, k=-5, D = 40, F = 5, find T. 

If 5= 38, r = A, fc«-5, D = 50, P = 4, find 5. 

If 5 = 150, r= |, * = 60000, Z) = 60, P = 5, find A:. 

If 5 = 200, r-1, 5 = 70000, &=-75,F = 5, find Z). 

If 5= 40, T=i, 5 = 65000, fc=-8, Z) = 120,findF. 

12. The horse-power of an electric current is given by 

EI 
H.P. = =j~, where E is the electromotive force and 7 the 

current in amperes. 
Solve for E and 7. 

If E =110, 7 = 30, find H.P. 

If H.P. = 6, E = 200, find /. 

If H.P. = 10, 7 = 40, findE. 

13. The heat generated by a current is given by H = • 24 EI T 
(Joule's Law) where H is the heat in calories, E the electro- 
motive force, I the current in amperes, T the time in seconds. 

Solve for E, I and T. 

If £ = 110, 1=2, T= 30, find H. 

If # = 500, E= 6, I = 10, find Z\ 

If # = 1000, 7 = • 5, T = 160, find £. 

14. The space traversed by a body starting from rest and 
moving with a uniform velocity is given by s = vt, where s is 
the space, v the velocity and t the time. 

Solve for v and t. 

If v = 12 ft. per sec, < = 25 sec, find s. 
If 5 = 300 ft., t = 15 sec, find v. 

If 5 = 500 ft., v = 20 ft. per sec, find t. 

15. The width of a single belt to transmit a given horse- 
power is given by W= — D ^ „ — , where W is the width of 

the belt in in., H the horse-power transmitted, P the allowable 
pull per in. of width of belt, S the speed of the belt in feet 
per min. 



FORMULAS 107 

Solve for H , P and S. 

If W = 13, P = 30, S = 3000, find W. 

IiW= S, P = 40, S=3500, find H. 

UW= 6, # = 20, S = 3200, find P. 

UW= 6, H = 22, P= 30, findS. 

16. The brake horse-power ' of an engine is given by 

0_ p D \7 

B. H.P.= , where B.H.P. is the brake horse-power, P 

ooUUU 

the reading of the scale beam, P the length of the arm 
in ft., N the revolutions per min. 
Solve for P, R and N. 

If P =180 lb., P= 48 in., N = 250, find B. H. P. 

IfP.ff.P.=30, R= 54 in., iV=120, find P. 

IfP.#.P. = 32, P = 2001b., #=180, find P. 

IfP.ff.P.=36, P = 2201b., iJ = 5 ft., find N. 



CHAPTER X. 
MENSURATION OF AREAS. 

85. Mensuration is that part of Mathematics which deals 
with the length of lines, the areas of surfaces, and the volumes 
of solids. 

86. To Find the Area of a Rectangle or of a Square. 

In Figure 27, ABCD is 
a rectangle, i.e., a quadri- 
lateral with its opposite 
sides parallel and its angles 
right angles. If we divide 
each side into inches and 
join as above we see by 
actually counting the small 
squares that the area of 
the rectangle is six square 
inches (to scale). This 
result might have been obtained by multiplying the number 
of inches in the length (3) by the number of inches in the 
width (2). From this example we infer a formula for the area 
of a rectangle. If A represents the area, b the length, 
and h the breadth, then A=bh, or the area of a rectangle 
= length X breadth. 

Note. — A correct statement of the above formula would 
manifestly be — the measure of the area of the rectangle = the 
measure of the length multiplied by the measure of the breadth, 
but for the sake of brevity the word " measure " will be 
omitted throughout. 

Make drawings in your laboratory book to test the accuracy 
of the above. 

108 



D 



Fig. 27 



MENSURATION OF AREAS 
87. To Find the Area of a Parallelogram. 



109 




Fig. 28 



In Figure 28, ABCD is a parallelogram (|| gm ), i.e., a quadri- 
lateral with its opposite sides parallel. 

If the right-angled triangle DFC be cut out and placed on 
EBA, it will coincide with EBA. The ||s m ABCD is therefore 
equal in area to the rectangle EBCF. If the area of the || gm 
is A, the base b, and the perpendicular height or altitude h, 
then A = bh, or the area of a parallelogram = base X perpen- 
dicular height. 

Make drawings in your laboratory book to test the accuracy 
of the above. 

88. To Find the Area of a Triangle in Terms of its Base and 
Altitude. 




Fig. 29 



In Figure 29, ABC is a triangle. If we draw CD parallel 
to AB and AD parallel to BC, we have the ||« m ABCD. Since 



no 



MATHEMATICS FOR TECHNICAL SCHOOLS 



the area of the || gm A BCD is bisected by its diagonal AC, we 
have the area of the triangle ABC as one-half the area of the 
|| gm ABCD. If A is the area of the triangle, b its base, and 
h its altitude, then A=%bh, or the area of a triangle = \ base 
X altitude. 

Make drawings in your laboratory book to test the accuracy 
of the above. 

89. To Find the Area of a Triangle in Terms of the Sides. 

In Figure 30 we have a 
triangle ABC and have de- 
noted the sides by a, b, c; a 
being opposite angle A, b 
opposite angle B, and c 
opposite angle C. 

The area of the triangle 
is given by the formula: 
A = -\/s(s-a) (s-b) (s-c) where 
a, b, c are the sides, and s 
is one-half their sum. 
Example: — If the sides in Figure 30 are 13 ft v 14 ft., 15 ft. 
respectively, then 5 = 21, s-a = 8, s-b = 7, s-c = Q. 
.*. .4 = V2lX8X7X6 = \/7056 = 84sq. ft. 
Exercises LVIII. 
1. Supply the missing quantities in the following rectangles : 

Area Length Breadth 




sq. ft. 


4 ft. 


3 ft. 


444 sq. ft. 


37 ft. 


ft. 


360-5 sq.ft. 


ft. 


18 -9 ft. 


sq. yd. 


24 ft. 9 in. 


15 ft. 6 in. 


f acre 


ft. 


2\ chains 



MENSURATION OF AREAS 



111 



2. Supply the missing quantities in the following parallelo- 
grams: 



Area 



Base 



Altitude 



sq. ft. 


4 ft. 


3 ft. 


48400 sq. yd. 


352 yd. 


yd. 


sq. ft. 


2 ft, 3 in. 


8 in. 


378 sq. in. 


3 ft. 6 in. 


in. 


sq. yds. 


5 yd. 1 ft. 


3 yd. 2 ft. 



90. To Find the Area of a Trapezium. 




Fig. 31 



Figure 31 represents a trapezium, i.e., a quadrilateral with a 
pair of sides parallel. 

The diagonal AC divides the trapezium into the two triangles 
ABC and ADC. 

Area of ABC = \ yh. 
Area of ADC = \ xh. 
/. A = \ yh+%xh = \h(x+y), 

or the area of a trapezium = sum of parallel sides X h the per- 
pendicular distance between them. 



112 



MATHEMATICS FOR TECHNICAL SCHOOLS 



91. A Practical Application of the Triangle and the Trapezium 
is found in the Measurement of Land. 

The following represents an entry in a surveyor's field book 
and the corresponding plan: 

Plan 





Field Book 




Links 




To B 




460 


to E 120 


340 




180 


to D80 


100 


From 


A 



90 to C 



go North. 



The field book entry is read upwards, 
which in the case above indicates that 
the chain line runs north from A. The 
centre column refers to measurement 
from A along the chain line to points 
from which the offsets are taken. Offsets to 
indicated on the right and offsets to the left 
on the left. 

In the plan the area of ADX = §X100X80 

the area of EDXZ=\ X240( 120 +80) « 

the area of BEZ = ^X120X120 

the area of A YC =1X180X90 

the area of B YC = §X280X90 




the right are 
are indicated 

= 4000 sq. li. 
= 24000 sq. li. 
= 7200 sq. li. 
= 8100 sq. li. 
= 12600 sq. li. 



Total Area =55900 sq. li. 
= -559 acres. 



MENSURATION OF AREAS 

Exercises LIX. 
1. Find the missing quantities in the following triangles: 



113 



Area 



Base 



Altitude 



sq. in. 


24 in. 


13 in. 


sq. ft. 


2 ft. 6 in. 


3 ft, 4 in. 


120 sq. in. 


in. 


15 in. 


22 sq. ft. 


5 ft, 6 in. 


ft. 


5§ acres 


320 rods 


yards 



2. Find the missing quantities in the dimensions of the fol- 
lowing boiler plates in the form of trapeziums: 



Area 



sq. in 


6 ft. 


62 in. 


102 in. 


sq. in. 


6 ft. 9 in. 


77 in. 


83 in. 


8505 sq. in. 


11 ft. 10 in. 


101 in. 


in. 


sq. in. 


8 ft. 5 in. 


79£ in. 


93 in. 


9841f sq. in. 


98| in. 


90| in. 




549 sq. ft. 


25 ft. 




18 ft. 



3. Find the areas of the following triangles: 

Sides 3 ft., 4 ft., 5 ft.; answer in sq. feet. 

Sides 4 yd., 2 ft., 3 yd., 2 ft., 1 yd., 1 ft.; answer in 
square yards. 

Sides 17 in., 18 in., 19 in.; answer in sq. inches. 



114 MATHEMATICS FOR TECHNICAL SCHOOLS 

Exercises LX. 

1. Find the areas of the triangular faces of a number of the 
models in the laboratory, using both methods. Make drawings 
in your laboratory book. 

2. Find the areas of trapeziums available in the laboratory. 
Make drawings in your laboratory book. 

3. A rhombus is a quadrilateral with all its sides equal. 
Construct a rhombus in your laboratory book having each side 
2 in. Employ both experiment and equality of triangles to 
establish how one diagonal divides the other, and also the 
magnitude of the angle contained by the diagonals. Write 
out the details and derive a formula for the area of a rhombus 
in terms of the diagonals. 

4. Take a series of measurements in the school grounds 
and enter in your laboratory book as suggested. Draw 
a plan to scale from your measurements and calculate the 
area. 

5. What is the area of the surface of a boiler plate 3' 8" 
by 1' 6"? 

6. How many square pieces of zinc 6" X 6" can be cut from a 
zinc plate 3' X 6'? 

7. What is the value of copper in an open copper tank 
measuring 4f " long, 3|" wide and 2\" deep; copper weighing 
12 lb. per sq. ft. and costing 40c per lb.? (No allowance being 
made for laps, seams or waste). 

8. The diagonals of a sheet of zinc in the form of a rhombus 
are 24" and 16". Find the area of the sheet. 

9. If a sheet of copper 5' X 10' weighs 500 lb., what is the 
weight per sq. ft.? 

10. How many sq. ft. of sheet copper will be required to 
make an open rectangular tank 7' long, 3' wide, and \\' deep, 
allowing 12% extra for waste? 

11. Find the cost of shingling the roof in the diagram on 
page 62 with shingles laid 4§ in. to the weather if material 
and labour cost $14 a square, of shingles, the eaves projecting 
2' (equivalent to a roof 34'X28' on plan). 

12. Find the cost of putting a slate roof on the building in 
the diagram on page 62, gauge 85", at $30 a square. 



MENSURATION OF AREAS 



115 




Fig. 33 



13. Find the cost of covering with 1" square sheeting the 
gable ends of the building in the diagram on page 61, width 
20', rise 10', material to cost $50 per M, allowing 8% for 
waste. 

14. Find the cost of cover- 
ing with 1" square sheeting 
the gable ends of the roof re- 
presented in Fig. 33 at $52 per 
M, allowing 10% for waste. 

15. In the map of a district 
it is found, that two of its 
boundaries are approximately 
parallel and equal to 13 miles 
and 18 miles. If the breadth 
is 8 miles find the area. 

16. In the quadrilateral ABCD the diagonal AC is 62" 
long, and the perpendiculars on AC from B and D are 15" 
and 12" respectively, find the area of the quadrilateral in 
sq. ft. 

17. Construct an equilateral triangle 2" to the side. Find 
its altitude. 

18. Construct an equilateral triangle with the length of 
the side taken at random. Denote it by x and find the altitude. 
What is the relation between the altitude and half the base? 

19. Find the area of a regular hexagon, if one of the sides be 
3". (Divide into six equilateral triangles). 

20. Find the side of an equilateral triangle equal in area to a 
triangle with sides 13", 14" and 15" respectively. 

21. How many sq. ft. are there in the surface of a board 18' 
long, 6" wide at one end and 14" wide at the other? 

22. How many sq. in. are there in a triangular plate, if one 
of its sides be 15", and the perpendicular on it from the opposite 
vertex be 8"? 

23. If the sides of a triangular plot of ground be 26', 28' 
and 30' respectively, find the length of the perpendicular from 
the opposite vertex on the 30 ft. side. 

24. One side of a triangular plate, containing 45 sq. in., is 
8". Find the length of the perpendicular on this side from the 
opposite vertex. 



116 



MATHEMATICS FOR TECHNICAL SCHOOLS 



25. The sides of a right-angled triangle are 5", 12" and 13" 
respectively. Find the areas of the equilateral triangles 
described on its sides. Do these areas bear any relation to 
each other? 



26. A column having a 



opposite arms of the cross 4 2 



cross-shaped section has two 
long, and the other two 



arms 4". The arms are \" wide. What is the area of the 
section? 

27. A T-shaped section has the top flange 8" long and f " 
wide, the other flange measuring 4' long by f " wide. What 
is the area of the T1 

28. The two parallel sides of a trapezium measure 13 chains 
60 links, and 6 chains 40 links; the other sides are equal, each 
being 8 chains 50 links. Find the area. 

29. ABCD is a quadrilateral in which the following measure- 
ments have been taken: AB = S0", BC = 17", CD = 25", 
DA =28", the diagonal BD = 2Q". Find the area in sq. ft. 

30. ABCD is a quadrilateral in which the angles ABC, 
CD A are right angles, and AB = 36 chains, BC = 77 chains, 
CD = 68 chains. Find the area in acres. 

31. Find the area of a quadrilateral ABCD in which the 
diagonal AC measures 30', and the perpendiculars on it from 
B and D are 3^' and 6' respectively. 

32. Draw the plan and calculate the area, in acres, of a plot 
of ground from the following* notes: 



Links 



• 


To B 






530 




to £75 


400 






240 


120 to C 


:o D 100 


150 




From 


A 


go North. 



MENSURATION OF AREAS 



117 



33. Draw a plan and calculate the area, in acres, from 
the following notes: 



Chains 





to B 






24-5 




ToF 2 


15-26 






10-1 


3-16 to E 


To D24 


8-6 ' 






4-3 


1-5 to C 


From 


A 


go North. 



34. Draw a plan and calculate the area, in acres, from the 
following notes: 



Links 





to B 








1200 






250 


100 








760 




324 





400 






50 


360 




200 


From 


A 


Go 


N. 30° W 



35. How many 6 in. sq. tiles should be supplied to cover 
the courtyard shown in Figure 34, an allowance of 5% being 
added to cover cutting and breakage? 



118 



MATHEMATICS FOR TECHNICAL SCHOOLS 



36. Figure 35 shows a gusset-plate for a girder. What is its 
weight if the plate is of mild steel \" thick, weighing 20-4 lb. 
per sq. ft.? 



t-/fl- 




WlU'^9 



U74-//VJ 




Fig. 34 



Fig. 35 



37. Calculate the length of the rafters on each pitch and the 
total area of the entire gable end of the building in Figure 36. 




■xf- 



Fig. 36 



-/5* — 

Fig. 37 




38. Determine the area of the cross-section in Figure 37. 




Fig. 38 



92. The Circle. A circle is a plane figure bounded by a line 
called the circumference and such that every point on it is 
equidistant from the centre. 



MENSURATION OF AREAS 



119 



93. To Find the Circumference of a Circle, the following 
measurements of a series of circular models were made and 
tabulated as follows: 



Circumference 


Diameter 


Circumference 

— ^ = w 

Diameter 


11-78 


in. 


3-75 


in. 


3-1413 


6-54 


in. 


2-08 


in. 


3.1442 


. 10-98 


in. 


3-5 


in. 


3.1371 


6-35 


in. 


2-02 


in. 


3 • 1435 


16-85 


in. 


5-36 


in. 


3-1436 


5-15 


in. 


1-64 


in. 


3 • 1402 


30-0 


in. 


9-54 


in. 


3 • 1446 


10-71 


in. 


3-41 


in. 


3 • 1407 


8-64 


in. 


2-75 


in. 


3-1418 




Average 


3-1418 



The value of tt has been determined to a great number of 
decimal places but 3-1416 is a close approximation. Since the 
fraction ^- =3-142 when carried out to the third place, it is 
commonly used as the value of tt. • 

From the above experiment we infer that Circumference = 
it Diameter or C = tZ). If in the formula C = n D we substitute 
for diameter its value in terms of the radius, we obtain 
C = 7r(2r) or C = 2:rr. 



120 



MATHEMATICS FOR TECHNICAL SCHOOLS 



94. To Find the Area of a Circle. If we take a circular board 
and divide it into sections as shown in Figures 39, 40, and place 
them as in Figure 41, we practically have a rectangle whose 
length is one-half the circumference and whose width is one- 
half the diameter of the circle. 





Fig. 39 



Fig. 40 




Fig. 41 

Hence, to find the area of a circle we multiply one-half the 
circumference by the radius, i.e., Trrxr = irr 2 . 

.*. A = irr 2 . 

In Figure 41, how would you to some extent overcome the 

difficulty of the length of the rectangle not being a straight line? 

In the formula A = irr 2 if we write for r its value in terms of 

D we get A „(|)'-^-?J^ = . 7854Z) , 

This formula for the area of a circle is commonly used by 
engineers and machinists. 



MENSURATION OF AREAS 



121 



Exercises LXI. 
1. Supply the missing quantities in the following circles: 

(tt = 3-1416) 



Radius 


Diameter 


Circumference 


Area 


5 ft. 










14 ft. 










3-1416 ft, 










150 sq. ft. 



95. To Find the Area of a Circular Ring or Annulus. 

The area of the outer circle is nR 2 
and the area of the inner circle is 
?rr 2 . .'. area of the ring = nR 2 — 
7rr 2 = 7r ( j R2_ r 2) =1F (R+r)(R-r). 

If we examine this latter formula 
in relation to the figure we see that 
U + r is the length of the mean 
diameter AB, and that R — r is the 
width of the ring. The formula 
therefore, be 




v{R+r)(R — r) may, 




Fig. 43 



circle by the arc ADB, 



Fig. 42 

written 7r(mean diameter) X (width 
of ring). .*. area of ring = mean 
circumference X width of ring. 

96. To Find the Length of the Arc 
of a Circle. 

In Figure 43 the chord AB divides 
the circumference of the circle into 
two arcs ADB and AHB. 

If the angle AOB, that is the 
angle subtended at the centre of the 
is 120° then the length of the arc 



122 MATHEMATICS FOR TECHNICAL SCHOOLS 

ADB is ^f# X circumference or in general the length of the 

n 
arc ADB is ^r^X circumference, where n is the number of 
3b0 

degrees in the angle subtended at the centre. 

The length of an arc may be found approximately by the 

formula: — Length of arc = — 5 — where c is the chord of the 

arc and a is the chord of half the arc. 

97. To Find the Area of a Sector of a Circle. A sector of a 
circle is that part contained by two radii and the arc cut off 
by them. 

In Figure 43, KOH represents a sector of a circle. If the 

angle K H be 60° the area of the sector will be -££$ of the 

n 
area of the circle, or in general the area of the sector is ttfk X 

00U 

area of the circle, where n is the number of degrees in the angle 

contained by the two radii. 

By a method similar to that used in finding the area of a 
circle it may be shown that the area of the sector =% arc of 
sector X radius of circle. 

98. To Find the Area of the Segment of a Circle. A segment 
of a circle is that part of the circle contained by an arc and its 
chord. 

In Figure 43 the chord A B divides the area of the circle into 
two segments, the area above AB being called the minor 
segment, and the area below AB the major segment. It is 
evident that the area of the segment ADB is equal to the 
area of the sector AOB minus the area of the triangle AOB, 
so that if sufficient data is given the area may be found by this 
method. The area of the minor segment in Figure 43 may be 
found approximately from the formula. 

Area of Segment = y +§ ch. Where c is the length of the 
chord AB and h is the height CD. 



MENSURATION OF AREAS 123 

Exercises LXII. 

1. Measure a number of circular objects as suggested on 
page 119. Tabulate in your laboratory book, find the ratio 
of the circumference to the diameter in each case, and take 
the average of these results. 

2. What length of steel sheet would be needed to roll into a 
drum 42" in diameter? 

3. What is the circumference in feet of an 18" emery wheel? 

4. The 72 in. drivers on a locomotive make 245 turns per 
min. How many feet will the locomotive go per min.? How 
many miles will it travel per hour? 

5. Find the diameter of a driving wheel measuring 15' 
8^" around the outside? 

6. If a point on the rim of a fly-wheel should not travel over 
a mile a minute, what should be the maximum speed, in 
revolutions per min., of a fly-wheel 7' in diameter? 

7. A pulley 3' in diameter makes 200 revolutions per min. 
Through how many feet does a point on its rim travel in 2 
minutes? 

8. If a belt connects the pulley in the preceding question 
to a 15" pulley, through ho.w many feet will a point on the 
rim of the latter travel in 1 minute? 

9. If a speed of a mile a minute is desired, what size emery 
wheel should be ordered to go on a spindle running 1320 
R.P.M.? 

10. A degree of latitude in Toronto measures 264613-31 ft. 
Find in miles the length of the parallel of latitude passing 
through Toronto. 

11. The perimeter of a semi-circle is 72", find its radius. 

12. A pulley 36" in diameter drives another pulley 14" 
in diameter. The belt velocity is 22' per second. What are 
the R.P.M. of the pulleys? 

13. The fly-wheel of an engine is 12' 6" in diameter and 
revolves at 96 R.P.M. It is belted to a 48" pulley on the main 
line shaft. Find the speed of the shaft. 

14. A 36" pulley making 143 R.P.M. is belted to another 
making 396 R.P.M. Find the diameter of the latter. 

15. Describe a circle of given radius on your cross-section 
paper. Count the squares as accurately as possible. Con- 
struct a square on the radius and count the squares. Divide 



124 



MATHEMATICS FOR TECHNICAL SCHOOLS 



the former result by the latter. Repeat this experiment 
changing the radius in each case. Tabulate in your laboratory 
book and find the average of your results. 

16. To establish A = -7854Z) 2 experimentally, cut a circular 
piece of cardboard 1' in diameter and also a square piece of 
the same material 1' to the side. Weigh both and find the 

value of — r^j— — j — r^— j — . Repeat this with pieces of board, 
weight of circle 

pieces of zinc, etc., taking care that the materials, in any one 

case, have the same thickness and density, and that the 

diameter of the circular part is the same as the side of the 

square. Tabulate in your laboratory book and find the 

average of the results. 

17. Fill in the omitted entries in the following: 



No. 


Diameter 


Circumference 


Area 


1 


7 






2 




44 




3 




• 


154 


4 




176 




5 






201| 


6 






55-44 


7 


14-8 






8 




264 




9 


15-4 







18. The piston of a locomotive is 20" in diameter. Find its 
area in sq. in. If the highest pressure carried is 205 lb. per 
sq. in., what would be the total pressure tending to blow off 
the cylinder head? 

19. A workman finds the circumference of a shaft to be 11". 
In order to find the strength of the shaft he must know the 
area of a cross-section. Find this area. 



MENSURATION OF AREAS 125 

20. Which has the greater capacity, one 4" pipe or two 
2" pipes? 

21. The area of an 8" circle is how many times the area of a 
4" circle? The area of a 12" circle is how many times the area 
of a 4" circle? 

22. Employ the relation between the sides of a right-angled 
triangle to find the diameter of a pipe equal in carrying capacity 
to two pipes 2" and 3" in diameter respectively. Illustrate 
by means of a diagram. Extend this method of illustration 
to find the diameter of a pipe equal in carrying capacity to 
three pipes 2", 3", 4" respectively, in diameter. 

23. The total pressure in a cylinder is to be 6000 lb. If 
the pressure per sq. in. is 50 lb., what is the diameter of the 
piston? 

24. A circular duct in a heating system is to supply air 
for four rectangular outlets 6" by 8". What must be the 
diameter of the duct so that its capacity will be equal to the 
combined capacity of the four outlets? 

25. Establish experimentally the formula — Area of ring = 
mean circumference X width — by considering the ring as a 
trapezium. 

26. The inner and outer diameters of a ring are 9" and 
10" respectively, find the area of the ring. 

27. A hollow cast-iron column has inside and outside 
diameters of 12" and 16" respectively, find the area of the 
end of the pipe. 

28. What is the area of a circular race track 378 yd. inside 
diameter and 16' wide? 

29. What is the area of the end of a cast-iron pipe that is 
12" outside diameter and 1" thick? 

30. What is the area of the end of a rod that is 4|" outside 
diameter, and has a \\" hole running through the centre 
of it? 

31. A circular court 150 yd. in diameter is to have a walk 
10' wide around it on the inside. The remainder is to be 
sodded. Find the totalcost if the walk costs $2.00 a sq. yd. 
and the sodding 40c a sq. yd. 

32. The radius of a circle is 8'. Find the area of a sector 
of the circle, the angle of which is 36°. 



126 



MATHEMATICS FOR TECHNICAL SCHOOLS 



33. Find the radius of a circle such that the area of a sector 
whose angle is 60° may be 182-5 sq. in. 

34. Find the area of the sector of a boiler supported by a 
gusset-stay, the radius of the boiler being 42" and the length 
of the arc 25". 

35. The centres of two circles which intersect aTe 12' apart. 
The radius of the one circle is 9', and that of the other 8'; 
find the area of the part which is common to both circles. 

36. Find the area of the segment of a circle if the chord 
be 15" long and the height of the arc 6". 

37. Construct an arc of a circle by tracing part way around 
any circular object. Join the ends of the arc to form the 
segment of a circle. Find the centre of the circle and determine 
the length of the arc by treating as part of the total circum- 
ference. Also find the length of the arc by the formula 

— r — , and hence determine the percentage error in this 

formula. 

38. Find the area of the segment in the above by finding 

the asea of the sector and subtracting the triangle. Also, 

h 3 
find area by the formula — +f ch, and hence determine the 

JZG 

percentage error in this formula. 

99. The Ellipse. An ellipse is a plane figure bounded by 
a curved line, such that the sum of the distances of any point 

in the bounding line from 
two fixed points is con- 
stant. Each of these fixed 
points is called a focus 
(plural foci). 

Figure 44 shows an el- 
lipse for which the sum 
of the distances of the 
point P from the foci F 
and F' is equal to the 
sum of the distances of 
any other point in the bounding line from F and F'. AB is 
called the major axis and CD the minor axis. 




Fig. 41 



MENSURATION OF AREAS 127 

To Construct the Ellipse. Place the given diameters AB 
and CD at right angles to each other at their centres E. 
From D with radius AE cut the major axis at F and F'. This 
gives the foci. Place pins at F and F' and also at D. Place 
a string around the three pins forming a triangle of string 
FDF'. Take out the pin at D and, substituting a pencil, 
trace as in Figure 44. If we represent the major axis by 2a 
and the minor axis by 26 the circumference of the ellipse is 
given by. the formula: 

Circumference = tt (a -\-b) (approximately) . 

The area of the ellipse is given by the formula Area = ■*■ ab. 

100. Regular Polygons. A regular polygon is a figure 
having all its sides equal and all its angles equal. 




Fig. 45 

ABCDE in Figure 45 is a regular pentagon (5 sides). 

If we bisect the angles A and E, the point of intersection 
of the bisectors will give us the centre of the inscribed 
circle of the pentagon, and hence a point equidistant from 
all the sides. This perpendicular from the centre on the side 
is called the apothem. The area of the triangle AOE = \ar, 
.*. area of pentagon = 5 X \ ar. 

Generally, then, the area of the polygon with n sides is 
\nar. 



128 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Since it is necessary to employ Trigonometry to find the 
apothem, the following table is given: 



Number of 
Sides. 


When Side is a 
Multiply a- by 


When Area is a 2 
Multiply a by 


3 


•433013 


1-519671 


4 


1 


1- 


5 


1-720477 


•762387 


6 


2-598076 


•620403 


7 


3-633912 


•524581 


8 


4-828427 


•455090 


9 


6-181823 


•402200 


10 


7 • 694209 


•360511 


11 


9-365640 


•326762 


12 


11-196150 


• 298858 


15 


17-642360 


• 238079 


18 


25-520770 


• 197949 


20 


31-567876 


• 177980 



Explanation of table. The first column gives the number* 
of sides, the second gives the area when the side is known, 
the third gives the side when the area is known. 

Thus, if the side of a five-sided regular polygon is 6 in. 
then the area is obtained by multiplying 6 2 by 1-720477; 
also if the area of a ten-sided regular polygon (decagon) is 
256 sq. in. the length of a side is obtained by multiplying 
•v/256 by -360511. 

Example: — The side of a twelve-sided regular polygon is 7". 
Find the area. 



MENSURATION OF AREAS 



129 



From the second column of the table — 

Area = 7X7XlM96150 = 548-61135 sq. in. 
Example: — The area of a nine-sided regular figure is 726 
sq. ft. Find the length of a side. 

From the third column the length of the side = V726 X • 402200 

= 10-8353'. 

101. Irregular Figures. Simpson's Rule for Finding Area. 

The area of an irregular figure may be accurately determined 
by the use of a planimeter, a description of which is given on 
page 131. When great accuracy is not required, a sufficiently ac- 
curate measurement may be made by the use of Simpson's Rule. 




Fig. 46 



Figure 46 represents an irregular figure 
divided into eight equal 
parts. The perpendiculars 
to this base line, d lf d 2 , d 3 , 
a 4 , a 5 , a 6 , a 7 , a 8 , a 9 , are 
called ordinates, and since 
there is an even number of 
divisions there will be an 
odd number of ordinates. 
The rule applies only when 
there is an odd number of 
ordinates. 

Consider Figure 47 consist- 
ing of the first two sections 
of Figure 46. 

HM is drawn through E 
parallel to BC and such that BK = KN = NC. 



The base line is 
o 




136 MATHEMATICS FOR TECHNICAL SCHOOLS 

The area of ABCD = area of A BKH +area of HKNM + 
area of MNCD approximately. 

Area of AB KH = $s(d l +dJ 
Area of HKNM=%sd 2 
Area of M N C D = ±s(d 2 +d 3 ) 
Area of ABCD = \ s(d x +±d 2 +d 3 ) 
Similarly the area of the third and fourth sections 

Similarly the area of the fifth and sixth sections 

= i*(d 5 +4d 6 +d 7 ) 
Similarly the area of the seventh and eighth sections 

= £s(d 7 +4d 8 +d 9 ) 
Adding these we get the total area 

-| s {(d 1 +d 9 )+W 2 +d i +d a +d 8 )+2(d a +d & +d 7 )} 

= i*C4+4JS+2C) 

Where A = sum of first and last ordinates. 
B = sum of the even ordinates. 
C = sum of the odd ordinates, omitting the first and 

last. 
s = common interval. 

102. The Planimeter. The name of the instrument comes 
from "planus" meaning flat, and "meter" meaning measure. 
As the principle of recording area is the same in both of the 
types shown, Figures 48, 49, we will confine our description to 
the compensating planimeter. Its use consists in tracing the 
contour of the figure to be measured with the tracer / as 
shown. When doing so the wheel M is made to revolve, and 
it is by the extent of these revolutions that the area of the 
traced figure is ascertained. The various parts of the planimeter 
are so dimensioned as to bring about one complete revolution 
of the wheel when an area of 10 sq. in. has been traversed. 



MENSURATION OF AREAS 



131 



Attached to the wheel is a white drum, divided into 100 
parts, one of which indicates an area of • 10 sq. in. By means 




r Hi. HJ- 



~A.MSI.KR PlaNIMETEH 




Fig. 49— Otts Compensating Planimeter 



of the vernier (see page 173) the single parts of the drum can 
be read to tenths, giving an area of -01 sq. in. To keep a 
record of the number of revolutions of the wheel a counting 
disc Z is attached to it by means of a worm-gear. Each mark 
on the disc corresponds to one revolution of the wheel, there- 
fore 10 revolutions of the wheel corresponds to one revolution 
of the disc. 



132 MATHEMATICS FOR TECHNICAL SCHOOLS 

Applying this to the reading in Figure 50 we have— the last 

number on the disc is 3, .'. 30 sq. 
in.; the last number on the drum is 
5, .'. 5 sq. in.; the last division 




2 



^ between the 5 and 6 is 8, .". -8 sq. 
^///////7?T^rf////////////////////Z. in.; the 4th division of the vernier 
FlG * m is opposite a division on the drum, 

.". 04 sq. in. Total reading 35-84 sq. in. 

Exercises LXIII. 

1. Construct an ellipse in your laboratory book as suggested. 
Write a note as to why your construction fulfils the require- 
ments. Find the area by counting the squares and check by 
formula for area. 

2. Construct an ellipse on cardboard having major and 
minor axes 4" and 2" respectively, and also a rectangle having 
length 4" and breadth 2". Cut out both, weigh, and find 

the value of — '- — ^ — ^ — — . Do the same with different 
wt. of ellipse 

materials and find the average of your results. 

3. A plot of ground in the form of an ellipse has major 
and minor axes, 200' and 150' respectively. Draw to scale 
in your laboratory book and find the perimeter and area. 

4. An elliptic man-hole door has major and minor axes of 
3' and 2' respectively. It is made of cast iron \" thick. 
Find weight if 1 cu. in. weighs -26 lb. 

5. At what distance from the end of the major axis should 
the hole for the centre of revolution be drilled in an elliptic 
gear whose axes are \\" and 2"? (Elliptic gears will mesh 
when revolving about their foci). 

6. The area of an elliptic sheet of zinc is 88 sq. in. If its 
minor axis is 4", find its major axis. 

7. The head of a hexagonal bolt is \" to the side; find the 
area of the head. 

8. A square is 4" to the side. An octagon is formed by 
cutting off the corners of the square. Find the side of the 
octagon and hence its area. Find the area by subtracting 
the areas of the four corners from the square and compare 
with previous result. 



MENSURATION OF AREAS 



133 



9. Ten hurdles, each 4' long, are placed to form a regular 
decagon. Find the area enclosed. 

10. A steel plate in the form of a regular pentagon measures 
If " on each side and is \" thick. Find its weight, if a cu. in. 
of steel weighs -283 lb. 

11. The area of a regular hexagon is 284-112 sq. in. Find 
a side of the hexagon. 

12. Regular polygons of 6 sides are inscribed in and circum- 
scribed about a circle of radius 1\ Find the difference of 
their areas. 

13. Construct a semicircle 4" in diameter in your laboratory 
book. Find its area by Simpson's rule. Check by means of 
formula for the area of a circle and thus calculate the per- 
centage error in Simpson's rule. 

14. Construct an ellipse with major and minor axes 4" and 
2" respectively. Proceed as in the preceding question. 

15. Make a drawing of Figure 
51 in your laboratory book. Com- 
mon interval f". If the scale 
be -£%" to the foot, find the area 
in square ft. by Simpson's rule. 
Check by planimeter and estimate 
percentage error. Note — areas of 
similar figures are to one another 
as the squares on corresponding 
sides. 

16. The ordinates of a curved 
piece of sheet lead in inches are 
20, 30, 29-9, 29-5, 28-4, 25-7, 14-2. The common distance 
between them is 3-65"; find the area. 

17. The half-ordinates of a transverse section of a vessel 
are in feet 12-2, 12-2, 121, 11-8, 11-2, 10, 7-3 respectively. 
The common interval is 18"; find the area. 



w- 


f 


p 


1< 


}> 


# 


p 


WS L 










-^ 


"A 










J J * * p. iy 




/'V/' Jf .o- 










Fk 


3. 51 









CHAPTER XI. 
RATIO AND PROPORTION. 

103. Ratio. We are constantly comparing weights, dis- 
tances, sizes, etc. If one piece of metal weighs 50 lb. and 
another 10 lb., we say that the first is five times as heavy as 
the second, or that the second is one-fifth as heavy as the 
first. If one board is 8 ft. long and another 2 ft. long, we 
say that the first is four times as long as the second, or that 
the second is one-fourth the length of the first. 

This Relation between Two Quantities of the same Kind is 
called Ratio. 

Note. — In the above definition of ratio it is important to 
notice "of the same kind." It would clearly be absurd to 
compare bushels and feet. 

A ratio may be written in two different ways. For example, 
the ratio of the diameters of two wheels which are 10 in. 
and 16 in. in diameter can be written as a fraction -ff. Again, 
since a fraction indicates division, i.e., 10-r-16, the line in 
the division sign is sometimes left out and the ratio is written 
10 : 16. In either case the ratio is read "as ten is to sixteen." 

Since a ratio may be expressed as a fraction it may be 
reduced to lower terms without changing its value. For 
example, if one casting weigh 600 lb. and another 150 lb., the 
ratio of the weight of the first to the weight of the second 

id 600-4 

1S T5~ff — T- 

Example 1: 

The diameter of the cylinder on an engine is 18" and the 
diameter of the piston rod is 3". What is the ratio of the 
cylinder diameter to the piston rod diameter? 
Diameter of cylinder 1 8 _ 

Diameter of piston rod ~ s ~ 1- 
134 



RATIO AND PROPORTION 135 

Example 2: 

A concrete mixture is made of cement, sand, and gravel in 

the ratio of 1 : 2\ : 5. If 25 bags of cement be used (1 bag 

= 1 cu. ft.), how many cu. ft. of sand and gravel will be 

required? 

25 (cu. ft. cement) X2§ = 62£ cu. ft. sand. 

25 (cu. ft. cement) X 5 =125 cu. ft. gravel. 

104. Proportion. When two Ratios are Equal, the Four 
Terms are said to be in Proportion. 

The two ratios 3 : 9 and 12 : 36 are evidently equal, since we 
can reduce 12 : 36 to 3 : 9. 

When written 3 : 9 = 12 : 36, these numbers form a proportion. 

Further we observe in the above proportion that the pro- 
duct of 3 and 36, i.e., the first and last, is equal to the product 
of 9 and 12, i.e., the second and third. 

The first and last are called the extremes, and the second 
and third are called the means. We, therefore, have : — The 
product of the means is equal to the product of the extremes. 

This relation may be expressed generally. 

If a, b, c and d represent the four terms of any proportion 
then: 

,, extremes N 

a : b = c : d 

^-means-^ 

Then in accordance with the above ad = bc. 

Application of this principle to some practical problems. 

Example 1: 

If it requires 60 men to turn out 200 shells in a day, how 
many men will be required to turn out 360 shells in a day? 

If x be the required number of men, then 60 : x = 200 : 360 

or 200z = 60X360 
60X360 
X ~ 200 ~ 108 ' 



136 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Example 2: 

If the diameter of a pulley is 40", and it makes 120R.P.M., 
what is the R.P.M. of a second pulley belted to the first if 
its diameter is 16"? 

Note. — When two pulleys are belted together, the larger 
of the two is the one that makes the least R.P.M. The pro- 
portion formed from their diameters and revolutions is, 
therefore, called an inverse proportion. 
Letx = R.P.M. of the second pulley, then 40: 16 = .r:120 

or 16x = 40X120 
40X120 



or s = 



16 



= 300 



After some practice in proportion we might write this 
directly— R.P.M. of 16 in. pulley = f£ of 120 = 300. 





105. Proportion in Similar Triangles : 

In triangle ABC and DEF, BC and EF are \\" and 3" 
respectively, also Z B = Z E and ZC= Z.F. 

The triangles ABC and DEF are, therefore, equiangular 
and are called similar triangles. If we compare corresponding 
sides with dividers we observe that DE = 2AB and DF = 2AC. 



RATIO AND PROPORTION 



137 



This experiment would suggest that, when triangles are 
equiangular, their corresponding sides are proportional. 

BC = AB = AC 
' EF DE DF' 
Make drawings in your laboratory book to verify the 
above. 

Observe this principle in the following Example : 




Fig. 54 

The top of a telegraph pole, Figure 54, is sighted across a 
5' pole placed 100' from the foot of the telegraph pole, the 
observer sighting from the ground at a distance of 15' from 
the foot of the 5' pole. Find the height of the telegraph pole. 
The triangles ABE and CDE are equiangular and therefore 
similar. 

CD = DE 
'*■ AB~BE 
CD 115 
5 " 15 

or, CD=^X^=38*'. 

Exercises LXIV. 

1. A room is 16' by 12'. What is the ratio of the length 
to the breadth? 

2. Two gear-wheels have 100 teeth and 40 teeth respectively. 
What is the ratio of the number of teeth? 

3. Detroit has a population of 1,000,000 and Toronto 
600,000. What is the ratio of the population of Toronto to 
that of Detroit? 



or, 



138 MATHEMATICS FOR TECHNICAL SCHOOLS 

4. A man rode 280 miles partly by rail and partly by boat. 
What distance did he travel by each, if the ratio is as 3 to 2? 

5. A locomotive has a heating surface of 1340 sq. ft. and 
a grate area of 24 sq. ft. What is the ratio of the heating 
surface to the grate area? 

6. The steam pressure in a locomotive is 196 lb. and the 
mean effective pressure in the cylinders is found to be 80 lb. 
What is the ratio of the mean effective pressure to the boiler 
pressure? 

7. Two pulleys connected together have diameters of 36" 
and 22". If the first makes 120 R.P.M., what is the R.P.M. 
of the second? 

8. The ratio of two gears connected together is as 5 to 3. 
The first makes 105 R.P.M. , how many does the second gear 
make? 

9. If 15 tons of coal cost $120, what will 18 tons cost? 

10. If the freight charges on a shipment be $40.50 for 
216 miles, what should it be for 300 miles? 

11. A pump discharges 20 gal. per min., and fills a tank 
in 24 hrs. How long would it take to fill the tank with a 
pump discharging 42 gal. per min.? 

12. A machinist gets $7.50 a day and a helper $5.00 a day. 
How much would the helper receive when the machinist gets 
$80.00, providing they both work the same number of days? 

13. If a yard-stick held upright casts a shadow 3' 9" long, 
how long a shadow would be cast at the same time by a 
chimney 66' 8" high? Check by drawing to scale. 

14. What is the height of a signal pole whose shadow is 
12', when a 10' pole at the same time casts a shadow of 2' 
4"? Check by drawing to scale. 

15. The length of shadow of a telegraph pole to the first 
cross-arm is 24'. A 6' pole at the same time casts a shadow 
of 2' 8". What is the height of the first cross-arm? Check 
by drawing to scale. 

16. In a mixture of copper, lead and tin there are 4 parts 
copper, 3 parts lead and 1 part tin. How many lb. of each 
would there be in 248 lb. of the mixture? 

17. A solder is made of 5 parts zinc, 2 parts tin, and 1 part 
lead. How many parts of each metal in 96 lb. of the 
mixture ? 



RATIO AND PROPORTION 139 

18. A man 6' high, standing 8' from a lamp-post, observes 
his shadow to be 6' in length. Find the height of a boy who 
easts a shadow of 4' when he stands 9' from the lamp-post. 

19. What will be the diameter of a gear that is to make 
60 R.P.M., if it is to mesh with a gear 24" in diameter, which 
makes 72 R.P.M.? 

20. What is the percent, grade of a road bed that rises 
1-2' in a horizontal distance of 40'? 

21. The roof of a house rises 1| # in a run of 2'. How far 
will it rise in a run of 20'? 

22. A road bed rises 2\' in 200', what is the percent, grade? 
In how many feet will it rise 1'? 

23. A cable railway up the side of a mountain has at places 
on the route a grade of 20%. What rise does this represent 
for every mile of horizontal distance? 



CHAPTER XII. 
SIMULTANEOUS EQUATIONS. 

Formulas — (continued). 

106. In the Discussion of the Simple Equation, we learned 
that it contained only one unknown quantity and that, 
therefore, the value of the unknown could be definitely de- 
termined. 

Thus, if 3.r+4 = 16 

3* = 12 
x = 4. 
If, however, a single equation contains two unknown 
quantities, we cannot find a definite value for either of them, 
and are restricted to finding the value of one in terms of the 
other. 

Thus, in the equation x-\-2y = 13, if we transpose 2y we 
have £ = 13 — 2y. The value of x in this equation will depend 
upon the values assigned to y. 

If, y = l, x = 13- 2 = 11. 
y = 2, x = 13- 4= 9. 
2/ = 5, a; = 13-10= 3. 

Similarly, if we give values to x we may obtain corresponding 
values for y. It is evident, then, that we cannot find definite 
values for x and y but have merely a statement of relation 
between them. 

If, however, another relation between x and y be obtained 

from the same problem, we can, from the two equations, 

determine the definite values of x and y. If we also find 

that 3z+2/ = 14, then transposing and dividing by 3 we have 

U-y 

T = 

x 3 . 

140 



SIMULTANEOUS EQUATIONS 141 



If, as above, y = l, x = 



14-1 



1 3 
3 



O 14 ~ 2 A 

y = 2, x=-^—=4:. 

14-5 
y = o, x = —^—=3. 

Comparing the two sets of values for x and y we observe 
that there is only one pair of values that will satisfy both equa- 
tions, namely, x = S and y = 5. These values of x and y for 
the two equations are called simultaneous values and the 
equations are known as simultaneous equations. 

107. In the following Problem we will illustrate Two Methods 
of Solution. One alloy contains 4% copper and another 
10% copper. How many pounds of each should be used to 
make a 100 pound mixture containing 6% copper? 

Solution by Substitution: 

Let # = No. of lb. from the first alloy. 
And ?/ = No. of lb. from the second alloy. 
Then x+y = 100. (1) 

4x 1 Oy 6 
also, Tnrf+TQA - = 7oo~ xl00 > which reduces to 4x + 10?/ = 600 

or, 2x + 5?/ = 300. (2) 

From (1) a: =100 -y 

Substituting in (2) 

2 (100-i/) +% = 300 

200-2?/ +5y = 300 

giving y = 33f. 

Substituting in (1) a: = 66f. 

Therefore we must use 66f lb. from the first alloy and 
33J lb. from the second. 



142 MATHEMATICS FOR TECHNICAL SCHOOLS 

Solution by Elimination : 

x+ y = 100 (1) 
2z+5?/ = 300 (2) 
(l)X2 = 2x+2?/ = 200 
(2) = 2x+5y = Z00 
Subtracting -3y = - 100 

2/ = 33i 
and x = 100-?/ = 66§. 
In simultaneous equations it frequently happens that one 
of the unknowns can be readily expressed in terms of the 
other, and the solution obtained by means of the simple 
equation. The problem given above affords an example of this. 
It has been observed that, if the equation contains only one 
unknown, the value of this unknown may be found from one 
equation; also that, if the equation contains two unknowns, 
two equations are necessary to find the values of the unknowns. 
This may be extended to three or more, and we say, 
generally, that we must have as many distinct equations 
as there are unknowns to be found. 

Exercises LXV. 
Solve the following equations: 

1. 3*+4y = 10. * ,* « 
4x + y = 9. 7 - 5 + 2~ 5 ' 

2. 4x+7?/ = 29. x-y = ±. 

*+3y=U. *a.»-ijB 

3. 8s- y = U. 8^3 

x+8y = 53. *_y = A 

4. 2x+oy = 25. 4 5 
3x- y = 0. 2 _ 

5. 3x-5y = 6. 9 ' x +y = 1 ' 

2x X * 

6. ^+ y = 16. 



10. f-!=0. 



J "4~"" 3x-£y = 17. 



■+I-M 



SIMULTANEOUS EQUATIONS 143 

Exercises LXVI. 

1. If scrap-iron contains 4% silicon and we have a pig-iron 
containing 9% silicon, how many pounds of each must be 
used to make a ton of mixture containing 6% silicon? 

2. The law of a machine is given by R = aE-\-b and it is 
found that when fl=100, £ = 25, and when R = 250, £ = 60, 
find a and b. 

3. One mixture for casting contains 20% copper and 
another mixture, for the same purpose, contains 8% copper. 
How many pounds of each should be taken to make 200 
pounds of a mixture containing 12% copper? 

4. The distance between the centres of two parallel shafts 
is 8". It is required to connect these shafts by a pair of 
gears so that one shaft will turn twice as fast as the other. 
Calculate the diameters of the gears. 

5. In a pulley-block lifting tackle, a force of 15 lb. will 
lift a load of 100 lb., and a force of 35 lb. will lift a load of 
300 lb. If the force (P lb.) and the load (W lb.) are related 
by an equation of the form P = mW-\-k find the values of m 
and k, and hence the law of the machine. 

6. If E represents the fixed expenses of a manufacturing 
company, V the variable expenses for each machine manu- 
factured, .V the number of machines per year, and C the 
total cost of operating, then C= E-\-VN. In 1918 the com- 
pany built 600 machines at a total cost of $18,000, and in 
1919, 800 machines at a total cost of $22,000. Calculate 
E and V. 

7. The law of a machine is given by E = aR-\-b and it is 
found that when ii = 10, £ = 5-46 and when R = 100, E = 
9-6; find a and b. 

8. The total cost C of a ship per hour is given by C = a-\-bs z 
where s is the speed in knots. When s is 10, C is found to be 
$26.00 and when s = 15, C is found to be $36.50. Find a and 
b and express the relation between C and s. 

9. At an election there were two candidates and 3478 
votes were cast. The successful candidate had a majority 
of 436. How many votes were cast for each? 

10. The moment of a force is the tendency of that force 
to produce rotation of a body, and is measured by the product 
of the force (in pounds) and the perpendicular distance (in 
feet) from the axis to the line of the applied force. For 



144 MATHEMATICS FOR TECHNICAL SCHOOLS 

equilibrium in the case of parallel forces, the algebraic sum 
of the forces is and the algebraic sum of the moments is 0. 

A uniform plank 20' long, weight 90 lb., rests on supports 
at its ends. A load of 500 lb. rests 8' from one end. Find 
the reactions of the supports. 

11. A uniform beam 16' long weighs 300 lb. It is supported 
at one end and at a point 4' from the other end. Calculate 
the reactions of the supports. 

12. A uniform beam, 12' long, is supported at each end and 
carries a distributed load, including its own weight, of \ ton 
per foot run. A concentrated load of 1 ton rests 5' from 
one end and another of 3 tons, 4' from the other end. Calculate 
the reactions of the supports. 

13. If ,f, =-34 and - = 1-47 find E and r. 

240+r r 

14. If Vi=V {l + Bt) and 1^ = 12-4 when * = 21-5 and 
^ = 17-3 when t = 7o-0, find V and B. 

15. The receipts of a railway company are divided as 
follows: — 40% for cost of operating; 10% for the reserve fund; 
a 6% dividend on the preferred stock which is j of the capital; 
and the remainder, $630,000, as dividend on the common 
stock, being at the rate of 4% per annum. Find the capital 
and receipts. 

Exercises LXVII. 

Formulas — (continued) . 

1. The time taken by a pendulum for a complete oscillation 

is given by 1 = 2tt\ - , where t is the time in seconds, I the 

length in ft. and g the acceleration due to gravity in ft. per 
sec. 

Solve for I and g: 

If / = ii, = 32-2 find t. 

If t = 2, g = 32-2 find I 

If * = l-57, 1 = 2 find g. 

2. The resultant of two forces at right angles is given by 
R= y/P 2 +Q i , where P and Q are the forces at right angles and 
R the resultant. 

Solve for P and Q: 

If P = 8, Q = 5, find R. 
If R = 17, Q = 8, find P. 



SIMULTANEOUS EQUATIONS 145 

3. The velocity of a body at the end of a specified time is 
given by v = u-\-at, where v is the final velocity in ft. per sec, 
u the initial velocity in ft. per sec, a the acceleration in ft. 
per sec per sec., t the time in seconds. 

Solve for v, u, a, t: 

Iiu = 12, a = lo,t= 6, find v. 
If v = 750, a = 30, t = 18, find u. 
If t' = 10-9, w = 45, < = 16, find a. 
Iff -215, u = 7o, a = 10, find*. 

4. The space traversed by a body is given by s= ut+%at 2 , 
where s, is the space in ft., u the initial velocity in ft. per 
sec, t the time in sec, a the acceleration in ft. per sec. 
per sec. 

Solve for u and a: 

Ifu = 20, a = 32-2, t= 8, find 5. 

If 5 = 300, a = 16, t= 5, find w. 

If s = 750, u = 25, t = 10, find a. 

5. The thickness of plate required in a boiler is given by 

'pd 
t = -^j , where t is the thickness in in., p the pressure in lb. per 



sq. in., d the diameter of the boiler in in., / the tensile stress in 


lb. per sq. in., 


e the efficiency of the joint. 




Solve for p, 


d, f, and e 






If p = 160, 


d = 8ft., 


/ = 20000, e = -7, 


find *. 


If t=-o, 


d= 90, 


/= 16000, e=-6, 


find p. 


Tf / — 3 

ii t — 8 , 


p = 150, 


/= 18000, e=-7, 


find d. 


If / = -*- 

A* «< — 16 > 


p = 140, 


e =-75, d = 72, 


find/. 


If < = A 

Al t — 16 , 


p = 120, 


d = 48, / = 5tons 


find e. 



6. The Kinetic energy of a falling body in foot-pounds is 

wv ^ 
given by K = -^— , where K is the energy, w the weight of the 

body in lb., i; the velocity in ft. per sec, g the acceleration due 
to gravity. 

Solve for iv, v, g: 

Ifw = 1000, v = 44, = 32-2, find K. 

li K=\12, w = 5000, y = 32-2, find v. 

IfX=12-4, t> = 35, = 32-2, find w. 



146 MATHEMATICS FOR TECHNICAL SCHOOLS 

7. The effort of friction (measured in lb.) in diminishing the 
load lifted is given by E=PV—W, where E is the effort of 
friction, P the effort in lb., W the weight in lb., and V the 

, ., ,. motion of effort 



motion of weight' 




Solve for P, V, W: 




IfP = 2-4, F=16, W = 25, 


find E. 


If E= 52, F = 16, P =4-2, 


find W 


l(E= 82, F = 16, W = Z0, 


find P. 


If £=104, P= 9, JF = 40, 


find V. 



8. The magnetic lines of force (Flux) is given by Q % = — s — , 

K 

where Q is the total flux, N the number of turns of wire in the 
coil, R the reluctance of the magnetic circuit, / the current in 
amperes. 

Solve for N, I, R: 

UN = 200, 7 = 5, fl=-0002, find Q. 

If Q = 30000, iV = 500, 7=15, find R. 

If Q = 100000, N = 50, R= -00005, find I. 

9. For a single riveted lap-joint, the efficiency in tension is 

P — d 
given by K t = — =— , where i£, is the efficiency in tension, P 

the pitch in in., d the diameter in in. of the rivet. 
Solve for P and d: 

If P=H in., d= f in., find K t . 
If #,= -5, d= A in., find P. 

If #,= -6, P=lf in., find d. 

10. The relation between a Centigrade and a Fahrenheit 
scale is given by C = f(P — 32), where C represents the Centi- 
grade and F the Fahrenheit reading. 

Solve for F: 

If F = 63°, find C. 
If C = 72°, find P. 
If C=-4°, find P. 

11. The counter electromotive-force (E.M.F.) of a motor is 
given by E= E C + IR, where E is the impressed E.M.F. , E c 
the counter E.M.F., /the current in amperes, R the resistance 
in the armature. 



R = 


•08, 


find E. 


R = 


•05, 


find E, 


I = 


25, 


find R 


R = 


•15, 


find /. 



SIMULTANEOUS EQUATIONS 147 

Solve for E c> I, R: 

If E c =108, I =40, 
If E =220, I =60, 
HE =110, E c = 108-5, 
If E =110, £ c = 109, 

12. The approximate length of an open belt connecting two 
wheels is given by L = %\{R-\-r)-\-2d, where L is the length, 
R and r the radii of the large and small wheels respectively, 
d the distance between the centres. 

Solve for R, r, d: 

If #=18 in., r = 10in., d = 40in., find L. 
If L= 15 ft., # = 16 in., r = 12in., find d. 

13. The approximate length of a crossed belt connecting two 
wheels is given by i = 3f (R-\-r) -\-2d, where L, R, r, and d 
have values as in question 12. 

Solve for R, r, d: 

If #=18 in., r=12in., d = 6ft., find L. 
IfZ = 12ft., # = 16in., r=8in., find d. 

14. The horse-power transmitted by belts is given by H.P. 

(T — T )V 
= qq nnn — » where 7\ is the tension on the tight side of the 

belt in lb., T 2 the tension on the slack side of the belt in lb. 
V the velocity in ft. per minute of the driver. 
Solve for T,-T 2 , V: 

If T l =120, r 2 = 50, F = 3141-6, find H.P. 

UH. P. = 81, V =2500, findr,-r 2 . 

15. The width of a single belt required to transmit a given 
horse-power at a given speed of the belt is expressed by 

W = — ' — , where W is the width in in., H.P. the horse- 
power, S the speed in ft. per min. 
Solveforff.P. and S: 

II H.P. = 100, S =3000, find IF. 

IfS =3200, IT = 6 in., findtf.P. 

Iftf.P.=40, IF = 5 in., find S. 

16. For a double belt the formula in question 15 becomes 
w = HJ\X?ti000 

SX100 ' 



148 MATHEMATICS FOR TECHNICAL SCHOOLS 



Solve for H. P. and S: 






If #.P. = 100, 


S =3000, 


find W. 


If 5 =3200, 


W = 6in., 


find H.P. 


If //.P. = 50, 


W = 5 in., 


find S. 



17. The length of belting in a closely rolled coil is given by 
i= -1309 N (D+d), where L is the length in ft., D the diameter 
of the roll in in., d the diameter of the eye in in., N the number 
of turns in the coil. 

Solve foriV, D, d: 

IiN= 15, D = 16h d = 5, find L. 

If L= 80, D = 14, d = 3, find N. 

If Z = 200, D = 44, iV = 30, find d. 

18. For a single riveted lap-joint the efficiency in shear is 

given by K s = p ' , where l£ s is the efficiency in shear, a 

the cross-section area of the rivet in sq. in., P the pitch of the 
rivet in in., T the thickness of the plate in in., S s the strength 
of rivet steel in shear (lb. per sq. in.), S t the strength of 
plate in tension (lb. per sq. in.). 

Solve for a, S s , P, T, S t : , 

If a= -7854, S s = 30000, P=Hin., 2T = f in., 

^ = 40000, find X s . 
IfX 4 = l-5, & = 32000, P = lfin., I 7 = 5 in., 
«S, = 35000, find a. 
-IfX, = l-75, a =-5, P = l-3 in., T = \ in., 

'«, = 38000, find S,. 

19. The horse-power of a boiler is given by B.H.P.= 

Q . K N/QA g 7 > where B.H.P. is the boiler horse-power, W the 

number of pounds of water evaporated per hour, H the total 
heat of steam above 32° F., t the temperature of the feed 
water. 



Solve for W, H,t: 








If W =20000, 


ff=1180, 


t = 100°, 


find B.H.P 


If B.H.P. =600, 


77 = 1175, 


t = 120°, 


find W. 


If B.H.P. = 650, 


H = 1200, 


W = 20000, 


find/. 



SIMULTANEOUS EQUATIONS 149 

20. The quality of steam (%) as determined by the throttling 

calorimeter is given by s = 100{ *L s -}, where x 

is the moisture in steam, H the total heat of steam at 
main pressure, h the total heat of saturated steam at pressure 
in calorimeter, T e the temperature of saturated steam at 
pressure in the calorimeter, T s the observed -temperature in- 
the calorimeter, C p the specific heat of superheated steam at 
constant pressure, L the latent heat of steam at main pressure. 

Solve for H, h, C p , T s , T e , L: 

Ifff=1180, A = 1150, r s = 220, T e = 215, C p =-48, 

Z = 920, finds. 
lix = ^{2%), A = 1160, 2^ = 225, r c = 218, 

C=-48, £ = 930, find H. 



CHAPTER XIII. 

GRAPHS. 

108. If we wish to fix the position of a point P on the page 
of this book, one way would be to find its perpendicular distance 
from the left of the page and also its perpendicular distance 
from the bottom of the page. If these distances were 3 in. 
and 4 in. respectively, then the point P would be definitely 
fixed with respect to the plane of the paper. 

Consider a sheet of paper ruled as in Figure 55. If we know 
that a point P is 6 divisions to the right of Y and 4 divisions 
above OX, we can, at once, locate the position of the point 
by counting 6 divisions along OX and then counting 4 divisions 
vertically to the point P. This method of fixing the point is 
called plotting the point. The lines OX and OY are called 
Axes of Reference, the point of intersection is called the 
Origin, and the distances 6 and 4, which locate the point, are 
called Co-ordinates. We would now say that the co-ordinates 
of P are 6 and 4, and would write it P (6, 4), the first number 
always giving the distance along OX and the second the distance 
along OY. OX is usually spoken of as the axis of X and 
Y as the axis of Y. Distances along OX are called abscissae 
and distances along Y are called ordinates. We see from the 
above that any point can be plotted on the squared paper if 
we know its distances from the axes Y and X. 

109. Let us use this for a Practical Purpose. A sewer runs 
across a rectangular lot and it is necessary to know its exact 
location in case of trouble later. 

150 



GRAPHS 



161 




A to cixv 
Fig. 55 



152 MATHEMATICS FOR TECHNICAL SCHOOLS 



_j uu|- ^4 1 , | i , £ :_i __ i , 

1!!==!=!==!=!===!; ' i ■ ■' :==: :' \^ 



Fig. 56 



GRAPHS 153 

Measurements are taken according to the following plan: 



Distance from Y 


5 


10 


15 


25 


35 


45 


55 


65 


75 


Distance from OX 





5 


10 


20 


30 


40 


50 


60 


70 



A graphical representation of the position of the sewer is 
shown in Figure 56. Each small division of the squared 
paper represents 1 ft. The first point A has for its co-ordinates 
(5, 0), the second point 5(10, 5), the third point C(15, 10) 
and so on. If we join these points we have a graph of the 
position of the sewer. At some subsequent date it is necessary 
to make an excavation for the footings of a building on this 
lot, and the contractor wishes to know if a particular footing 
will come too near the sewer. He takes measurements and 
finds that the distance from the side corresponding to O Y is 
30 ft., and the distance from the side corresponding to OX 
is 25 ft. While these distances are not actually recorded in 
the data previously taken, yet by going out 30 ft. (30 spaces) 
from OY and up 25 ft. (25 spaces) from OX, he would find 
that he is directly over the sewer. This illustration brings 
out one of the most important functions of a graph: It 
gives results for data not actually recorded at the outset. 

110. It usually happens in practice that we require to make 
a record of two corresponding sets of measurements, in which 
the unit of measurement in one is entirely different from the 
unit in the other. 

The following will illustrate: . 

The observations below were taken of the loads on a 

lighting plant from 3 P.M. to 12 P.M. at intervals of one 
hour. 



Time in hours. . . . 


3 


4 


5 


6 


7 


8 


9 


10 


11 
45 


12 


Load in Kilowatts. 


50 


60 


76 


120 


140 


150 


142 


100 


30 



164 



MATHEMATICS FOR TECHNICAL SCHOOLS 




S * 3 « 

Fig. 57 



GRAPHS 



155 



Figure 57 shows how the relation of time and load 
may be represented graphically. , Along the axis of X we let 
each main division represent one hour, while along the axis 
of Y we let each main division represent 20 kilowatts. The 
letters, A, B, C, D, E, F, G, H, I, J, represent the locations 
of the observations. By drawing a curve through these 
points, we have a graph which will show at a glance the varia- 
tions in load. We see, further, that it will give us the probable 
load for times in between those recorded in the data above. 

For example, if we wished to know the load at 8.30 P.M. we 
would take a point half-way between 8 and 9 on the axis of 
X and draw a perpendicular to it, represented by the heavy 
line in the figure. 

From the point where this meets the curve, draw a line 
perpendicular to the axis of Y as represented, and we have 
148 kilowatts as the probable load at 8.30 P.M. 

111. Frequently a Relation between Measurements is 
expressed by an Algebraic Equation. Suppose we wished to 
find the Fahrenheit reading corresponding to a Centigrade 
reading in degrees. Since 180 degrees Fahrenheit, measuring 
the range from 32° to 212°, are equal to 100 degrees Centigrade, 
measuring the range from 0° to 100° we have : 

100° Centigrade = 180° Fahrenheit. 

a° Centigrade = |£# a° Fahrenheit = | a° Fahrenheit. 

If 6° represent the Fahrenheit reading corresponding to a° 
Centigrade, then the relation is given by the equation b =fa+32. 

By giving different values to a in the equation we can obtain 
the corresponding values of b. These may be tabulated as 
follows : 



Values of a 



32 


10 

50 


20 


30 


40 


60 


80 


100 






Corresponding values of b 


68 


86 


104 


140 


176 


212 



156 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Figure 58 is a graphical representation of this algebraical 
relation. Along the axis of Y we have represented the values 
of a, while along the axis of X we have the corresponding 
values of b. Corresponding values other than those recorded 
may be read off from the graph. Thus 77° Fahrenheit equals 
25° Centigrade as represented in the drawing. 

Another value of graphs is here illustrated, that is, they 
act as checks on computations. 

112. It is often important to represent more than one Set of 
Relations on the same Sheet. 

The following are the results obtained with a wheel and 
axle mounted on ordinary plain bearings. W represents the 
load lifted in pounds, P the effort applied in pounds, F the 
friction measured in pounds, E the efficiency percent. 



w 


P 


F 


E 





•8 


1-60 





5 


4-30 


3-60 


58-2 


10 


7-14 


4-28 


70-1 


15 


9-91 


4-82 


75-7 


20 


12-81 


5-62 


78-0 


25 


15-63 


6-26 


800 


30 


18-50 


7-00 


81-2 


35 


21-50 


8-00 


81-4 


40 


24-45 


8-90 


81-8 



GRAPHS 



157 




Fig. 58 



158 MATHEMATICS FOR TECHNICAL SCHOOLS 



4+H 










I i 



■ '•" "i - — : — i — i — ; — 

ZO 25 »<> 

LOA.PVH *OUMP» 



Fig. 59 



In Figure 59, the lower line represents the relation between 
the load and the effort. The middle line represents the 
relation between the load and the friction. The top line 
represents the relation between the load and the efficiency 
percent. 

Note.— In the drawing of a graph relating to machines it 
often happens that the points are not absolutely on a straight 
line. It is necessary, in such a case, to take the line which 
lies most nearly along the path of the points. 



GRAPHS 



159 




(<SNOi. OOSI a3Ao)dlHC J — «NO\SIAla IITHS z 

Fig. 60 



113. Sometimes the sole Purpose of a Graph is to picture in a 
concise and striking way the relation between two measurements. 

Figure 60 is a graph (from a Toronto daily paper) based on 
the official weekly figures of losses sustained by the British 
merchant fleet during the height of the submarine warfare. 



160 



MATHEMATICS TOR TECHNICAL SCHOOLS 




Fig. 61 



GRAPHS 



161 




Fig. 62 



162 MATHEMATICS FOR TECHNICAL SCHOOLS 

114. In Business Transactions frequent use is made of the 
Pictograph. This kind of representation takes a variety of 
forms — varying sized men may represent populations, varying 
sized bales of cotton may represent the export of cotton, and 
so on. 

Figure 61, called the bar pictograph, is of frequent use. 

Example : 

The net income of a certain railway company for a recent 
year was divided as follows: 

Sinking Fund Requirements, $2,000,000. 
Dividend on Preferred Stock, $5,000,000. 
Dividend on Common Stock, $18,000,000. 
Additions and Improvements, $2,400,000. 
Surplus to Profit and Loss, $6,200,000. 

In Figure 61, the above amounts are represented by a series of 
parallel bars, each main division on the vertical line representing 
$2,000,000. When the division of the income was presented 
in this form, the directors saw at a glance the relative division 
of the returns from the road. 

This form of pictograph is also extensively used to represent 
a decline or growth in business. 

Example: — The graph on page 161 is from a report re 
Toronto's gross funded debt 1910-1919. 

It illustrates clearly the rapidity of the growth in recent 
years and its arrest in 1919. 

When it is necessary to represent a percentage division, the 
circular pictograph is of common use. 



GRAPHS 163 

Example. — The following figure is also from a report re 
Toronto's debt for 1919: 




MWJ .RfevEKlOE PRODUCVNG ceoSS TUMOEO DEBT 

' 1 Kok-RevekuE PRooocinq f> iai,8i3,7&3 

Fig. 63. 



The object of this figure is to show the percentage of the 
debt, which is due to each investment mentioned. 

Exercises LXVIII. 

Note. — Before attempting to draw a graph of any relation, it is 
important to make a careful study of the squared paper at 
our disposal for the work. The larger the graph the greater 
accuracy and range of readings; we should, therefore, draw 
our graph to cover, if possible, all the sheet. Further, we 
should so divide the horizontal and vertical distances that as 
many as possible of our readings may come exactly on the 
lines of the paper. 



164 



MATHEMATICS FOR TECHNICAL SCHOOLS 



1. The distances along a road from a certain point and the 
height of the road above sea-level at these distances are shown 
as follows: 



Distance from start- 
ing point in miles. 





1 


2 


3 


4 


5 


6 


7 


8 


Height above sea- 
level in feet 


60 


75 


90 


140 


175 


230 


260 


290 


330 



Represent the above relations by means of a graph. Esti- 
mate the probable height above sea-level 1\ miles from the 
starting point. 

2. The following table represents the output of an auto- 
mobile firm for the past ten years: 



Year 


1910 


1911 


1912 


1913 


1914 


1915 


1916 


1917 


1918 


1919 


Number. . . . 


700 


780 


850 


900 


1000 


950 


960 


1020 


1050 


1850 



Represent the above relations graphically. 

3. The following table gives the revolutions per minute of a 
60 in. diameter locomotive driver and the corresponding 
speed of the locomotive in miles per hour: 



Revolutions per 
min 






60 


90 


100 


150 


200 


250 


275 


300 






Miles per hour. . 


10-5 


15-7 


17-5 


26-3 


35 


39-3 


48-2 


52-5 



Represent the above graphically and find the revolutions 
for a speed of 30 miles an hour. 

4. The following observations of temperature were recorded 
on July 25, 1919: 



Hour of the day 


4 A.M. 


6 A.M. 


8 A.M. 


10 a.m. 


12 NOON 


2 P.M. 


4 P.M. 


6 P.M. 


8 P.M. 


10 P.M. 


Temperature in 


40 


45 


60 


70 


75 


85 


90 


80 


64 


60 



Draw a graphical representation of this variation in tem- 
perature. 

5. If a cu. in. of steel weighs • 28 lb. construct a graph showing 
relation between volumes and weights. 



GRAPHS 



165 



6. If 1 inch = 2-54 centimetres, construct a graph showing 
relation between the two systems of measurement. 

7. The following is an extract from a table giving breaking 
strength of steel, in pounds per sq. in., in relation to the per- 
centage of carbon in the steel: 





.09 


.18 


.20 


.31 


.39 


.50 


.57 


.71 


.79 






Breaking Strength 


53000 


64000 


05000 


77000 


90000 


97000 


110000 


124000 


127000 



Represent the above graphically and estimate the percentage 
carbon for a breaking strength of 100,000 pounds per sq. in. 

8. The prices charged by a manufacturing concern for a 
certain motor of different horse-powers is given by the following 
table : 



H.P 


1 


2 


3 


4 


5 


74 

' 2 


10 


15 






Price 


100 


140 


165 


180 


200 


250 


275 


325 



Represent graphically the relation between H.P. and price. 

9. The quotations of a certain industrial stock at intervals 
of a week, were, 48, 49, 52$, 53, 56|, 58, 56$, 55$, 53. 

Represent graphically the probable fluctuations in price. 

10. The record of a patient's temperature for a certain time 
at intervals of a half-hour, is 97, 97-5, 98, 98$, 99$, 101, 101$, 
102, 101, 100. Represent the fluctuations graphically. 

11. A tram-car is found to travel the distance y feet in x 
seconds, the distance moved in different times being measured 
and recorded as follows: 



Distance in feet 


(y)o 


7-5 


13 


20 


27 


34 


42 


49-5 


57-5 


Time in seconds 


(x)0 


1 


2 


3 


4 


5 


6 


7 


8 



Represent this relation graphically. 

12. A company finds that the buying expenses are 15% of 
its gross income; office expenses 5%; management 10%; other 
overhead 25%; selling expenses 30%; interest 10%; dividends 
3%; incidentals 2%. Use the pictograph to represent the 
division of the gross income. 



166 



MATHEMATICS FOR TECHNICAL SCHOOLS 



13. The value of the exports and imports of the United 
States for a given period is as follows: 



Year 


1830 


1840 


1850 


1860 


1870 


1880 


1890 


1900 


Value in 
Millions. . 


134 


222 


318 


687 


829 


1504 


1647 


2100 



Use the pictograph to represent this growth in commerce. 

14. The following results were obtained by hanging a 
series of weights on the free end of a spiral spring and thereby 
stretching it: 



Weight in lb. 






1 
•2 


2 

•4 


3 
•6 


4 

•8 


5 


6 


7 


8 


9 


10 


Stretch in in. 


10 


1-2 


1-4 


1-6 


1-8 


20 



Represent this relation graphically and indicate the probable 
stretch for a load of 5| lb. 

15. The following results were obtained as in the preceding 
except that the stretch in inches is given by differences: 



Stretch in in. 
by differences 



Plot the above in two parts — the first for loads up to 60 lb., 
the second for loads above. Compare the two graphs. 

16. A car starting from rest is drawn by a varying force F 
pounds, which, after t seconds, is as shown in the following 
table: 



t (seconds) 





2 


5 


8 


11 


13 


16 


19 


20 


F (pounds) 


1280 


1270 


1220 


1110 


905 


800 


720 


670 


660 



If the frictional resistance is constant and equal to 500 lb., 
draw a graph of the above relation and indicate the force after 
10 seconds. 

17. The elasticity of a wire may be found by twisting. The 
following readings were taken in experimenting with a steel 
wire: 



GRAPHS 



167 



Load in lb 





1 


2 


4 


5 


6 


Angle of twist in 
degrees 





6 


12 


24 


29 


26 



Represent graphically and indicate the probable twist for 
a load of 45 Tb. 

18. The law of a machine is given by the relation P = • 08 
FF+1-4. P being the force in pounds required to raise a 
weight W. The following values of W are given: — 21, 36-25, 
66-2, 87-5, 103-75, 120, 152-5. Find the corresponding 
values of P and plot the relation. Find the force necessary 
to raise a weight of 310 lb. from your graph. 

19. In a certain machine, P is the force in pounds required to 
raise a weight W. The following corresponding values of P 
and W were obtained experimentally: 



p 


2-8 


3-7 


4-8 


5-5 


6-5 


7-3 


8-0 


9-5 


10-4 


11-75 


w 


20-0 


25-0 


31-7 


35-6 


450 


52-4 


57-5 


650 


71-0 


82-5 



Draw the graph connecting P and W , and read the value of 
P when FT = 70. Also determine the law of the machine, and 
from it the weight that could be raised by a force of 45 pounds. 
(P=aW+b). 

20. The length of one degree on a parallel of latitude is given 
for certain latitudes as follows: 



Latitude. . . 





10° 


20° 


30° 


40° 


50° 


60° 


70° 


80° 


90° 


Length in 
Miles.. . . 


69-2 


68-1 


65 


60 


53 1 


44-6 


34-7 


23-7 


121 






Draw a graph of the above and employ it to estimate the 
length of a degree in latitudes 15°, 45°, 73°. 

21. The following are the results obtained with a set of 
rope pulleys: 








5.5 


12.2 


17.1 


2.5.0 


31.0 


37.5 


44.8 


50.8 


62.0 


76.0 


Effort in lb 


.94 


3 


5.5 


7.1 


10.2 


12.2 


14.5 


17.2 


19.7 


25.5 


30 








3 76 


6.5 


9.8 


11.3 


15.8 


17.8 


20.5 


24.0 


28.0 


40.0 


43 9 











45.8 


55.8 


60.2 


61.3 


63.5 


64.6 


65.2 


64.5 


60.8 


63.4 



168 



MATHEMATICS FOR TECHNICAL SCHOOLS 



On the same sheet of paper draw graphs of the relation 
between load and effort, load and friction, load and efficiency. 
From your graph estimate the effort necessary to lift a load of 
40 lb., also the friction and efficiency for this load. 

22. From a series of tests on an oil engine the following 
values of the weight of oil used per hour (W) and the Brake 
Horse Power (B.H.P.) were obtained: 



B.H.P 


10 


2-1 


30 


4-2 


4-70 


5-3 


W\b 


107 


216 


2-85 


3-91 


4-40 


4-90 



Represent the above graphically and estimate B.H.P. when 
JP«41b. 

23. Toronto required $30,080,000 during 1920, to meet 
civic expenses. This was obtained as follows: 

General taxes $13,074,312 

School taxes 6,396,788 

Water rates 2,840,066 

Surplus from 1919 2,415,345 

Hydro 606,069 

Local improvements 1,605,675 

Street railway 1,098,651 



Abattoir 

Rentals 

Licenses 

City car lines . . 

C.N.E 

Fines 

Other revenues. 



130,000 
186,600 
113,000 
445,000 
100,000 
150,000 
917,120 



Employ the circular pictograph to represent the above. 
24. The following are the results obtained with a screw-jack: 



Load in lb 





5 


10 


15 


20* 


28 


30 


35 


40 


45 


Effort in lb 


.172 


.282 


.359 


.409 


.578 


.688 


.797 


.960 


1.000 


1.100 


Friction in lb. . 


19.86 


27.48 






46.77 




62.04 




75.50 


83.13 


Efficiency %.. . 





15.4 






29.9 




32.6 




34.6 


35.1 



On the same sheet of paper draw graphs of the relation 
between load and effort, load and friction, load and efficiency. 
Estimate the missing quantities from your graph. 



GRAPHS 



169 



25. The results shown in the following table were obtained 
experimentally from a lifting machine. Plot the two curves 
connecting P and W and F and W: 



Load (W) lb 





5 


10 


15 


20 


25 


30 


35 


40 






Effort (P) lb 


.094 


.45 


.81 


1.17 


1.53 


1.88 




2.61 


2.97 






Friction (F) lb 


2.34 


6.32 


10.31 


14.29 


18.28 


22.26 






34.21 



Estimate the missing quantities from your graph. 

26. The tax rate in Toronto in 1919 was 28£ mills, divided 
as follows: 

General City purposes 10.89 mills 

Schools 7.90 mills 

Public Library 0.25 mills 

Administration of Justice 2.27 mills 

Street Maintenance 3 . 93 mills 

War Expenditure 3 . 26 mills 

Employ the circular pictograph to represent the above. 

27. The increase in wages of the employees of a railway 
company from 1913 to 1916, based on $1 a day, is given as 
follows : 

Trackmen from $1 . 15 to 

Station Agents from 1.75 to 

Office Clerks from 2. 10 to 

Trainmen from 1 . 80 to 

Machinists from 2 . 25 to 

Conductors from 3. 15 to 

Enginemen from 3 . 60 to 

Represent these increases graphically. 

28. The following table gives the edible portions of various 
kinds of fish and the price per pound: 



1.30 
2.25 
50 
80 
20 
25 



4.75 



Kind 


Halibut 


Haddock 


Whitefish 


Bass 


Herring 


Perch 


Pike 


Canned 
Salmon 


Edible portion 
in % 


72 


49 


56 


45 


57 


37 


42 


86 


Price per lb.... 


24 . 


18 


20 


22 


16 


12 


18 


32 



Express graphically the edible portions of these various kinds 
of fish that can be bought for $1. 



170 



MATHEMATICS FOR TECHNICAL SCHOOLS 



29. 



« a 



u 
























ji 


^ 


co 




E 


o 










© 






Ci 




CO 


- 






it 

c 






* 




'5. 

a 


o 


© 


G 

en 


co 


o 






00 






















s 














9 


5 


o 


© 


a 




t>. 






















■3 














* 


















J 


t~ 














■o 




J 


% 


3 


o 


o 
o 

8 


■ 


- 


o 


1 


CN 


5 








2 




o 


J 




o 

s 


a 


9 

0) 




m 


» 








o 




5 




3 




4 


~ 


a 


*. 




o 


00 






00 


00 






•<1< 






a 








a 


o 


# 




2 




00 




< 




7 






























o 








■a 






s 


•* 




j* 


oS 






a 






1 


C 






o 


g 




m 


t^ 




g 


<• 


00 


-H 










§ 




* 

o 


a 






t>. 




M 


O 




c 




iO 








CO 






§ 




00 




is 












* 


* 






o 








CO 






"5 


00 




rt 


b- 






H 


CO 


1 




















e- 


5 








o 








3 








CO 






















S3 


o 

00 


o 

00 


■ 




CN 


"i 












CN 




o 


< 






^9 






* 


-o 






o 


i 




o 


m 


B 




CO 


00 






CN 




a 


< 




1 


c^ 








-3 








O 




o 




o 


2 


co 

CN 




£ 










= 








« 


^ 


© 




5 


o 


— 

00 




CN 















CHAPTER XIV. 



MATHEMATICS OF THE MACHINE SHOP. 

115. Machinist's Scale. A machinist's scale is made of steel 
and is usually either 6" or 12" in length. There are markings 
on the four edges, 8ths, 16ths, 32nds, and 64ths. In measuring 
machinists prefer to " split ". a 32nd, instead of attempting 
to read to 64ths. 

116. Try Square. The try- 
square is used for testing if 
surfaces are at right angles 
to one another. The dia- 
gram illustrates its use for 
testing a piece of work. 

117. Calipers. Frequently 
it is not possible to obtain 
an accurate measurement 
with the scale, for example 
the outside diameter of a fig. 64 
cylinder, or the inside diameter of a pipe. For such purposes 
calipers are used. These are of three types — outside calipers, 
inside calipers, and hermaphrodite calipers. 







1 


BLADE 


o o 

o 
o o 

1 

<0 















172 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Outside calipers are used for taking outside dimensions as 
the diameter of a cylindrical piece of work, inside calipers for 
measurements such as the bore of a pipe, and hermaphrodite 
calipers for finding the centre of a piece of work and for 
scribing. 

The calipers must be finely adjusted so that they will just 
touch the sides of the work as they pass over it. Care must 
also be taken to keep them at right angles to the work. 




Fig. 66 

In laying the calipers on the scale to find the length, place 
one leg at the end of the scale and read the mark on the 
scale where the other leg touches (see Figure 66). 

118. Centring. Work is frequently held in a lathe between 
two points called centres. In order to accommodate these, 
small holes must be drilled in the ends of the work. These 
holes are countersunk to the same angle as the centres, 
usually 60°. 



MATHEMATICS OF THE MACHINE SHOP 



173 



(1) Centring by hermaphrodite calipers. The calipers are 

set so that the pointed leg reaches 
approximately the centre of the 
work. The calipers are then placed 
at A, B, C and D and arcs are 
described as shown. The centre of 
the work will be the centre of the 
figure thus obtained. 




Fig. (i7 



(2) Centring by the centre square. 
The centre square consists of a 
head and blade. The head is so 
adjusted that the edge of the blade 
comes across the diameter of a piece 
of round stock placed in the head as 
shown. 

A line is drawn along the blade on 
the work. The work is then turned 
to some other position and another 
diameter is drawn. Where these 
diameters cross will be the centre. 




119. Vernier. With the scale, 
Figure 69, we could measure to a cer- 
tain degree of accuracy. If the length Fia 
came between 7 and 8, we could estimate the amount, say 




174 



MATHEMATICS FOR TECHNICAL SCHOOLS 



7-6. For obtaining greater accuracy in this part between the 
7 and 8 a device known as a vernier is used (Pierre Vernier 
—1631). 




Fig. 69 



In Figure 70 a second scale CD, called a vernier, is placed 
alongside of the scale AB and we observe that 10 divisions 
on the vernier is equal to 9 divisions on the scale. Obviously 




each division on CD is T V less than a division on AB. There- 
fore the length between the 1 mark on AB and th.e 1 mark 
on CD will be T V of a division on A B. Also the length between 
the 2 mark on AB and the 2 mark on CD will be ^ of a 
division on AB, and so on. 




In Figure 71 the reading on AB is 2 plus a decimal. To 
get the decimal part we observe that division 4 of the vernier 
coincides with a division on the scale. Evidently the excess 
of the reading over 2 is the difference between 4 divisions 
on AB and 4 vernier divisions, which as above explained is 
T % of a division on AB. Therefore the reading is 2-4. 



MATHEMATICS OF THE MACHINE SHOP 



175 



For more accurate readings the vernier sometimes has 25 
divisions corresponding to 24 divisions on the scale. 

120. Micrometer. A micrometer is an instrument for 
measuring to a greater degree of accuracy than can be measured 
with a scale. 




Fig. 72 

The above is a representation of the micrometer for 
measuring in inches, the parts being indicated. 

The principle of the instrument is as follows: 

The screw is threaded inside of the sleeve with 40 threads 
to the inch. The thimble is attached to the end of the screw 
and the work to be measured is placed between the screw 
and anvil. The micrometer is then closed on the work by- 
turning up the thimble. 

Since the screw has 40 threads to the inch, one turn of the 
thimble closes the opening T ^th of an inch or -025". Each 
mark on the sleeve represents one complete turn of the thimble, 



therefore, four turns equals 4X T V" or 



i " 



Figures are placed 



on the sleeve at every fourth mark, representing tenths of 
an inch. The thimble is divided into 25 equal divisions so 
that turning the thimble one division advances the screw ^ 

of i =_JL_" 
U1 T¥ 10 • 

To read the micrometer in the above figure: 
The last number exposed on the sleeve is 4, therefore we 
set down T V or -4". Between the last number exposed and 



176 



MATHEMATICS FOR TECHNICAL SCHOOLS 



the edge of the thimble two small divisions are showing, 
therefore 2X T V" or 2X -025* = -05". 

The thimble has evidently turned 12 spaces from the zero 
mark, therefore |f of ^" = T ^"= -012". 

.-.total reading =• 4"+ -05"+ -012" =-462". 

If the micrometer has a vernier, of the type described, the 
divisions on the thimble can be divided, making the micrometer 
read to 10,000ths. 

Exercises LXIX. 



— & 



rrrrnT=: 



o I 2 b 



?r~ 



minimi 



*^*_ 



C « Z 3 A S 

Imlmliiiliiiliiili 



Fig. 73 

1. Calculate the micrometer readings in the above figures. 

2. If the thimble be turned backward through 6 complete 
revolutions, what decimal of an inch is the micrometer 
opened? 

3. Through how many turns must the thimble be moved 
to open the micrometer -7"? 

4. The sleeve reading is 4 and the thimble reading is 18. 
What is the opening of the micrometer? 

5. How many turns must the micrometer be opened to 
read -458"? 

6. A spindle is ground to 1-345". What is the setting on 
the micrometer? 

7. A ball measures -864". What is the setting of the 
micrometer? 

8. Calculate the setting of the micrometer for -^". 

9. Explain how you would set a micrometer for tAt," 
over |*. 

10. Calculate the setting of the micrometer for f ". 



MATHEMATICS OF THE MACHINE SHOP 



177 



121. Vernier Caliper. 

bar with a sliding jaw. 



The vernier caliper consists of a 
The bar is divided the same as the 



t y»*AAw«y 





1 

4-56769 

nlni iiilinlmlmliii 


nil 


mi mil iimnm 


O 5 fc> 15 20 2S 




* A 



\ 2 3f 



iMlMWlMiliHli. 



Fig. 74 




Fig. 74a 

sleeve of the micrometer, i.e., the smallest division being ^ 
of an inch. 

On the sliding-jaw is a vernier. It is divided into 25 parts, 
the total length of these parts being equal to 24 divisions on 
the bar. As previously described in the case of the vernier, 
the distance between say the 4th mark on the vernier and 
the 4th mark on the bar will be -^ of a division on the bar. 
Since a division on the bar is ^ of an inch, this distance will 
be T VXirff = TTnnr of an inch. 



178 MATHEMATICS FOR TECHNICAL SCHOOLS 

In the preceding figure our object is to obtain the bar reading 
opposite the on the vernier. The last figure showing on the 
bar is 3, .'. '3". From the 3 on the bar to the last division 
before the on the vernier we have three small divisions. 
.'. 3 X T V = 3 X -025" = -075". To get the vernier reading 
we observe that the 5 line on the vernier is exactly opposite a 
line on the bar, .\ -nftnr" = -005" 

.'. total reading = -3" + -075" + -005" = -380". 

Exercises LXX. 

1. What would be the correct setting for a vernier caliper 
to read 1-642"? 

2. A reading on the vernier caliper shows 1", 3 tenths, 2 
small divisions, while the 12th division on the vernier is in 
line with a beam division. What is the reading? 

3. How would you set a vernier caliper to read §"? 

4. What fraction of an inch is represented when the bar 
shows 2 tenths, 2 small divisions, and the 6th division on the 
vernier is in line with a beam division? 

5. How would you set a vernier caliper to read -7645"? 

6. A reading on the vernier caliper shows 3", 3 tenths, 3 
small divisions, while the 8th division on the vernier is in 
line with a beam division. What is the reading? 

122. Cutting and Surface Speed. In the running of 
machinery in the shop, the workman should know the speed 
at which to run the machine in order to give the best results. 
Lathes, milling machines, etc., are provided with attachments 
for changing the speed. This speed depends on the kind of 
material in the work, whether it is a roughing or finishing cut, etc. 

In the lathe the cutting speed is the rate at which the work 
passes the tool and is usually reckoned in feet per min. The 
same definition would apply to the cutting speed of a planer. 
In the shaper, however, it is the tool that is moving and as 
a consequence the cutting speed would be the rate at which the 
tool passes over the work. 



MATHEMATICS OF THE MACHINE SHOP 179 

Cutting Speed of a Lathe. If a piece of work is being 
turned in a lathe, the tool will pass over the whole circum- 
ference of the work in a complete revolution. If the diameter 
of the work be 7", then the circumference would be ^ X7" 
= 22"=2_§'. if the lathe is making 40 revolutions per min., 
then the cutting speed of the lathe = ff X 40 = *£* = 73^' per min. 

From the above we have that the cutting speed of the lathe 
in feet per minute = Circumference of work in ft.XR.P.M. 
(R.P.M. being contraction for "revolutions per minute"). 

We have here three quantities involved — the cutting speed, 
the circumference of the work, and the revolutions per minute. 
If we know any two of these we can find the third. 

Example 1: 

A piece of work 5" in diameter is to be turned in a lathe. 
How many revolutions per min. should the lathe make to 
give a cutting speed of 35 ft. per min.? 

The circumference of the work = 5 X ^ =ip."= j^l'. 

The tool travels -^- ' in 1 revolution. 

It would travel 35' in yyVX t 5 - revs. = 26+revs. 

Example 2: 

The surface speed of an emery-wheel is to be 200 ft. per 
min. It is belted to an arbor to run 50 R.P.M. Find the 
circumference of the wheel. 

In 50 revolutions a point on the surface travels 200'. 

In 1 revolution £££■ = 4'. 

.'. circumference of wheel = 4'. 
^-Z) = 48". 

U — 2^ Ay — AOyy . 

Exercises LXXI. 

1. A piece of steel f" in diameter is turned in a lathe at 
100 R.P.M. What is the cutting speed? 

2. A brass rod £" in diameter is revolving at the rate of 
300 R.P.M.; find the cutting speed. 



180 MATHEMATICS FOR TECHNICAL SCHOOLS 

3. A piece of work with a diameter of 3" is being turned 
at a cutting speed of 50 ft. per min. What are the R.P.M.? 

4. A 30" grinding wheel is run at 50 R.P.M. What is the 
surface speed? 

5. At what R.P.M. should a 50" wheel be run, for a surface 
speed of 300 ft. per min.? 

6. What sized wheel should be ordered to go on a spindle 
running 1600 R.P.M., to give a surface speed of 4000 ft. 
per min.? 

7. An 8" shaft is being run to give a cutting speed of 50 ft. 
per min. What are the R.P.M.? 

8. A cast-iron pulley is machined at a cutting speed of 30 
ft. per min. If the R.P.M. is 10, what is the diameter of the 
pulley? 

9. What would be the rim speed in ft. per min. of a fly- 
wheel 10' in diameter, running 75 R.P.M.? 

10. How many revolutions per min. will it take to turn a 
piece of tool steel 2" in diameter with a cutting speed of 40 
ft. per min.? 

123. Cutting Feed. In turning a piece of work in the 
lathe, the feed is the number of revolutions of the work to one 
inch travel of the carriage. 

In drilling, the feed is the number of revolutions necessary to 
cause the drill to descend 1 in. 

Example 1: 

How many revolutions are necessary to take one cut over 
a shaft 6' in length with a feed of 30? 

Length of shaft = 72". 

.*. number of revolutions = 72X30 = 2160. 

Example 2: 

How long will be necessary to take one cut over a shaft 3' 
long and 3" in diameter, with a cutting speed of 30 ft. per 
min. and a feed of 34? 

Circumference of work = 3X-^r- = -3r-" = ft' 

Since cutting speed is 30 

.'. R.P.M. = 30-h ff = 30Xff 



MATHEMATICS OF THE MACHINE SHOP 181 

Revs, necessary to finish the work = 36X34 

.'. time required = (36 X 34) h- (30 X ft) =32 -[-min. 

On account of variations in the nature of materials used, 
especially of cast-iron, and also in the cutting capacity of 
tool steels, no fixed rule can be given for cutting speeds 
and feeds. Generally, for roughing — slow speed and heavy 
feed; for finishing — high speed and light feed. 

Exercises LXXII. 

1. How many revolutions will be necessary to take a cut 
over a steel rod 8' in length with a feed of 24? 

2. How long will be necessary to take a cut over a shaft 
22" long and 2\" in diameter with a feed of 20 and a speed 
of 30 ft. per min.? 

3. A piece of work 5' in length is being turned at the rate 
of 60 R.P.M. If the feed be 16, what time will be necessary 
to make one complete cut? 

4. A cast-iron pulley is 18" in diameter and has a 6" face. 
If the cutting speed be 40 ft. per min. and the feed 16, how 
long will it take for one cut over the work? 

5. A shaft 6' long and 4" in diameter is being turned at a 
cutting speed of 30 ft. per min. If the feed is 20, what 
fraction of the surface will be cut over in 15 min.? 

6. A drill is being fed to the work at -01" per revolution. 
If it makes 40 revolutions per min., in what time will it cut 
through 2" of metal? 

7. A drill cuts \\" into a piece of work in 15 minutes. 
If it makes 36 revolutions per min., what is the feed of the 
drill? 

8. A drill with a feed of 100 is making 50 revolutions per 
min. In what time will it cut through 2\" of metal? 

9. In 10 min. one cut is taken over a shaft 3' long and 4" 
in diameter. If the feed of the machine is 21, what is the 
cutting speed? 

10. It takes 12 min. to take one cut over a shaft 18" long 
and 3" in diameter. If the cutting speed is 40 ft. per min., 
what is the feed? 



182 



MATHEMATICS FOR TECHNICAL SCHOOLS 



124. The Trigonometrical Ratios. It is frequently necessary 
to make use of trigonometrical ratios in the machine shop. 
We will merely define these ratios without giving reasons for 
the names assigned. 



1. The Sine of an angle 

2. The Cosine of an angle 

3. The Tangent of an angle 

4. The Cosecant of an angle = 

5. The Secant of an angle = 

6. The Cotangent of an angle = 



_Side Opposite 
Hypotenuse 
Side Adjacent 



Hypotenuse 
_Side Opposite 
Side Adjacent' 

Hypotenuse 
Side Opposite' 

Hypotenuse 
Side Adjacent' 
Side Adjacent 



Side Opposite' 

The contractions Sin, Cos, Tan, Cosec, Sec, Cot, are used 
when writing the above. 

A 




Fig. 75 



In the above triangle: 
a . ^ AC n . BC 
SmB = AB- CosB = AB- 

SecB = ^. CotB = B 



Tan B = 



AC 
BC 



Cosec B = -r-r, 
AC 



BC AC 

Tables giving the values of the trigonometrical ratios of 
all angles from 0° to 90° are available. 



MATHEMATICS OF THE MACHINE SHOP 



183 



125. Taper. The taper on a piece of conical work is the 
difference in diameter for one foot of the work. 



r 



IZ 



Fig. 76 



If the work in the figure be 12" in length and the diame- 
ters 2\" and 2", the difference of \" in the diameters is called 
the amount of taper, i.e., one-half inch per foot. 

D 




Fig. 77 

Taper expressed as an Angle. 

In the taper above the sides DA and CB when produced 
meet at E. The angle AEB is known as the angle of taper. 
Assume that the length H N of the piece is 12", that the 
diameter AB of the small end is 6", and the angle of taper 
I AEB is 10°. 

In A AMD : DM = 12 tan 5° 
= 1.04988" 

.'.DC = 6 +2X1 -04988 = 8-1 (Approx.) 

The amount of taper is therefore 2-1 in. per foot. 

Kinds of Tapers: 

(1) Morse Taper. 

Possibly the most common taper is the Morse. It is found 
chiefly on lathe spindles, drill spindles, and grinder spindles. 



184 



MATHEMATICS FOR TECHNICAL SCHOOLS 



It is approximately f " per ft., but varies somewhat according 
to the following table: 



Number 


Inches per Foot 





•625 


1 


•600 


2 


•602 


3 


•602 


4 


•623 


5 


•630 


6 


•626 



Of the above numbers 1, 2 and 3 are more commonly used. 

(2) Brown and Sharpe Taper (B. & S.). The Brown and 
Sharpe taper is \ in. per ft. for all sizes except No. 10, 
which is -516 in. per ft. 

It is the taper used on milling machine arbors, the milling 
machine having been developed largely by Brown and Sharpe. 

(3) Jarno Taper. The Jarno taper is -6 in. per ft. for all 
sizes. It is frequently used on lathe centres. 

126. Methods of Cutting Tapers on the Engine Lathe. 

(1) By Means of the Compound Rest. 

In cutting short tapers and bevels, this compound rest 
(Fig. 78) is used, the extent of the work being limited by the 
length of the compound rest screw. This attachment is used in 
turning head-stock centres. A graduated slide divided into 
degrees permits of adjustment to any required angle. 

(2) By Offsetting the Tail Stock. 

When the tail centre and head centre of the lathe are in 
alignment, the cutting tool moves in a line parallel to a line 



MATHEMATICS OF THE MACHINE SHOP 



185 



connecting the two centres. If a piece of work be turned 
in this position, a uniform cut will be taken throughout its 




Fig. 78 



length. If, however, the tail centre be moved out of align- 
ment with the head centre, the cut will be deeper at one end 
than at the other. 

The following diagram will help to make this clear: 



hEAD CENTKE 




1_TA\L CENTRE 



Fig. 79 

In the above diagram the tail centre is represented as set 
over an amount x. If a piece of work be turned when the 



186 MATHEMATICS FOR TECHNICAL SCHOOLS 

centres are related in this way, the radius of the work at the 
tail centre will be less by x than the radius of the work at the 
head centre. Since the taper is the difference in diameter 
between the centres, it follows that the offset of the tail stock 
is one-half this difference in diameter. 

Example : 

A piece of work 9" long is to be turned with a taper of \" 
per foot; find the amount of offset of the tail stock. 

A taper of \" in 12" = a taper of ^Xh" in 9". 
= a taper of f" in 9". 

As the tail stock must be set over one-half of this amount, 
the required offset is tV". 

In the above method it must be kept in mind that the 
amount of offset of the tail stock is one-half the difference of 
the end diameters whether the taper extends the full length 
of the work or not. 

Example : 

A steel pin 12" long is to be tapered for 8" and turned 
straight for the remaining 4". The diameter at the large end 
is If" and the small end is to be 1" in diameter; find the 
amount of offset. 

The taper in 8" = If " - l" = f " 

.-.taper in 12"=-^ off" = ||" 

.'.amount of offset = £§". 

(3) By Means of a Taper Attachment. 

Many lathes are now fitted with a taper attachment. This 
is attached to the back of the lathe and is connected to the 
cross-feed. A movable slide can be adjusted at various angles 
to the travel of the carriage, and the cross-feed screw having 
been released, the cross-feed slide will move backward and 
forward according to the alignment of the slide on the taper 
attachment. This method should always be used if a lathe 
with a taper attachment is available. 






MATHEMATICS OF THE MACHINE SHOP 187 

Exercises LXXIII. 

1. A piece of steel 8" long has end diameters of 1" and f\ 
Find the amount of taper, i.e., taper per foot. 

2. A piece of work 10" long has a Jarno taper and has a 
diameter at the large end of If". What is the diameter at 
the small end? 

3. A piece of work 6" long has a No. 1 Morse taper and has 
a diameter at the small end of tk". What is the diameter 
at the large end? 

4. A taper pin tapers \" per foot and has end diameters 
of \ w and f". What is the length of the pin? 

5. A piece of work with a Jarno taper has end diameters 
of A" and H". What is the length of the work? 

6. A piece of work with a B. & S. taper has end diameters 
of 1J* and 1^"- What is the length of the work? 

7. A piece of work 15" long has end diameters of \\" and 
2-gz". What kind of taper was used in turning? 

8. A piece of work 18" long has end diameters of \\" and 
2". What kind of taper was used in turning? 

9. A piece of work 2f" long has end diameters of -368" 
and -475"; find the amount of taper. 

10. A piece of work 6" long is to have end diameters of 
•45" and -7625". What kind of taper would the work have 
when finished? 

11. A piece of work 9" long is to have end diameters of 
•3125" and -6875". What is the amount of taper? 

12. A piece of work 21" long has end diameters of -875" 
and 1-925". What kind of taper has it? 

13. Explain why the offset of the tail stock is one half 
the difference of the required diameters. 

14. A steel pin 1\" long is to be turned with a taper of 
f " per foot. What is the offset of the tail stock? 

15. The diameter at the large end of a piece of work is \\" 
and the diameter at the small end 1". What is the offset of 
the tail stock? 

16. A taper pin is 1" in diameter at the large end and ^-" 
at the small end. What is the offset of the tail stock? 

17. A taper gauge has end diameters of 2\" and If". If 
the length of the taper is 9" and the total length 12", find the 
offset of the tail stock. 



188 



MATHEMATICS FOR TECHNICAL SCHOOLS 



18. Determine the distance that the tail stock should be 
set over to cut the following: 

(a) A No. Morse taper on a piece of work 9" long. 
(6) A No. 1 Morse taper on a piece of work 1\" long. 

(c) A Jarno taper on a piece of work 10" long. 

(d) A Brown & Sharpe taper on a piece of work 18" long. 

19. A piece of work 18" long is to be turned straight for 
12" and the remaining 6" to be tapered. The diameter at the 
large end is to be 2" and at the small end 1"; find the offset 
of the tail stock. 

20. A tapered piece of work is 8" long, and a Jarno taper 
was turned on the piece. What is the difference in end 
diameters? 

21. A piece of work 20" long is to be turned to a diameter 
of 3" at the centre, and to be tapered from centre to each 
end with a taper of \" per foot. Determine the end diameters 
and the offset of the tail stock. 

22. A piece of work 12" long having a diameter at the 
larger end of 6", tapers to an angle of 10°. What is the 
amount of taper? 

23. A piece of work 18" long and a diameter at the smaller 
end of 8", tapers to an angle of 8°. What is the amount of 
taper? 

24. What is the angle of taper in a Morse No. 0, a Morse 
No. 2, a B. & S., a Jarno? 

127. Threads. A thread is formed by cutting a uniform 
spiral groove around a piece of work. 

aaAA 






¥1_L 



The diameter of a screw is the distance from the point of 
a thread on one side to a point on the opposite side (outside 
diameter of diagram). The inside diameter is the diameter 
measured at the bottom of the groove (see diagram). The 
pitch of a thread on a screw is the distance from the middle 



MATHEMATICS OF THE MACHINE SHOP 189 

point of one thread to the middle point of the next, measured 
in a line parallel to the axis. Pitch is usually stated as the 
number of threads per inch. Thus if there are 10 threads 
per inch, the pitch is ^. 

Stated generally : Pitch = — 55 7—7 1 : — : 

A o. oj threads per inch. 

To estimate the number of threads per inch, place a mark 
on the scale on the point of a thread and count the number 
of grooves within the inch line, or count the number of threads 
and subtract 1. 

The lead of a screw is the distance the screw advances in 
one complete turn. In a single threaded screw the pitch is 
equal to the lead. Thus if the pitch is ^, the screw will move 
forward -jV' in one complete revolution. In a double threaded 
screw the pitch is \ the lead, in a triple threaded screw •§ the 
lead, and so on. 

If a screw has a right-handed thread it turns in the direction 
of the hands of a clock when screwed into the nut. If a left- 
handed thread it will turn in the opposite direction when 
screwed into the nut. 

Exercises LXXIV. 

1. Secure a number of different kinds of screws and find 
the number of threads per inch in each. 

2. What is the lead of a single threaded screw if it has 
(a) 6 threads per inch, (6) 12 threads per inch, (c)15 threads 
per inch? 

3. A single threaded screw advances 2" in 12 turns, what 
is the pitch? 

4. What is the pitch of a double threaded screw if it has 
12 threads per inch? 

5. A jack-screw has 4 threads per inch. How far does it 
move in \ a revolution? 

6. What is the pitch of a triple threaded screw that advances 
3" in 6 revolutions? 

7. What is the pitch of a double threaded screw which 
advances 1" in 6 revolutions? 



190 MATHEMATICS FOR TECHNICAL SCHOOLS 

128. Kinds of Threads: 
(1) Sharp "V" Thread. 




u 


1 


t\ ♦ 


< 


ll\ " 





1 5 


M 


1 * 




I 






11/ ? 




1/ * 





V 


1 




Fig. 81 

The sharp "V" thread is a thread having its sides at an 
angle of 60° to each other and being perfectly sharp at both 
top and bottom. It is difficult to get a sharp "V" thread 
on account of the wear on the point of the tool in cutting. 

Depth of " V " thread. 

In figure above the thread has a pitch of 1", then in the 
triangle ABC each side is 1" in length. The depth of the 
thread will be equal to BD, the altitude of the triangle. 

In the right-angled triangle BCD, DB 2 = BC 2 -CD 2 
.-. D£ 2 = l 2 -(i) 2 or Z># = V!=-866". 

If the pitch be only §*, then since the triangle formed would 
be similar to the triangle ABC of the preceding, the depth 
would be | of -866"= -433". If the pitch be ^" then for like 
reason the depth would be & of -866"= -0721". 

Calculations for threads are usually made on the double 
depth. In a thread of 1" pitch the double depth would be 
2X -866" = 1-732". 

Since by the above the depth is proportional to the pitch, 
1*732 is used as a constant for all "V" threads. 

Example : 

If the pitch of a "V" thread is ^" the double depth would 
be T V of 1-732= -1732". 

Since pitch = Number of threads per inch' 

. , blg , th= 1-732 _ f 

U P Number of threads per inch 

1- 732 

or, for brevity, D = — ^-i where D is the double depth, an( 

N the number of threads per inch. 



MATHEMATICS OF THE MACHINE SHOP 191 

As the root diameter of the thread is the same size as the 
hole to be bored for tapping the thread, it is necessary to be 
able to find this double depth in selecting the size of drill. 
Example: 

What sized tap drill must be used for \" screw, sharp "V" 
thread, having 12 threads per inch? 

Root Diameter = Outside Diameter — Double Depth 
1-732" 
~' 5 " 12 
= .5"_. 1443"=- 3557". 

From the table of decimal equivalents ff is the next above, 
therefore the correct size. 

Exercises LXXV. 

1. By means of the method used in the preceding find the 
double depth of sharp "V" threads having 8, 12, 14, 18, 20 
threads per inch. Check by formula. 

2. If the double depth of a sharp "V" thread is -1443", 
find the number of threads per inch. 

3. If the double depth of a sharp "V" thread is -1732", 
find the number of threads per inch. 

4. What size of tap drill would be necessary for a |" screw 
with a sharp "V" thread having 9 threads per inch? 

5. What size of tap drill would be necessary for a If" screw 
with a sharp "V" thread having 7 threads per inch? 

6. What size of tap drill would be necessary for a -fa" screw 
with a sharp "V" thread having 12 threads per inch? 

7. If the single depth of a sharp "V" thread is -1733", 
find the pitch. 

8. The root diameter of a 3" bolt with a sharp " V" thread 
is 2-5052". What is the number of threads per inch? 

9. The root diameter of a bolt with a sharp "V" thread is 
•7835". If it has 8 threads per inch, what is the outside 
diameter? 

10. What is the root diameter of a 2" bolt with a sharp 
"V" thread having 4| threads per inch? 



192 MATHEMATICS FOR TECHNICAL SCHOOLS 

(2) The United States' Standard Thread (U.S. Std.). 





Fig. 82 

This thread is commonly used in machine work as it gets 
over the difficulty of the sharp edges of the "V" thread. 

This thread has the same triangular form as the sharp 
"V" thread but is flattened at the point and bottom. This 
flattened part is | of the pitch in width. As f of the height 
is taken from the top and bottom the depth of the thread is 
f the depth of the "V" thread. .-. depth = f of -866"= -649", 
Also double depth is f of 1-732" = 1-299". 

As in the sharp "V" thread, the double depth of the U.S. 
Std. for different pitches may be found by dividing the con- 
stant by the number of threads per inch. 

1-299 
.'. Double Depth of U.S. Std. thread = 



Also Root Diameter = Outside Diameter— 



Number of threads per inch 
1-299 



Nu mber of threads per in ch 

To find the size of tap drill for a U.S. Std. thread we would 
proceed as in the case of a sharp "V" thread. 

Example : 

What sized tap drill would be used for a f " screw, U.S. Std. 
thread, 11 threads per inch? 

Root Diameter = Outside Diameter — Double Depth 

1 -299" 
= . 6 25"- -ff-=' 5069". 

From table of decimal equivalents ff is the next above and 
consequently the correct size. 



MATHEMATICS OF THE MACHINE SHOP 



193 



Exercises LXXVI. 

1. What is the double depth of a U.S. Std. thread of | pitch? 

2. What is the root diameter of a &" U.S. Std. threaded 
screw, 12 threads per inch? 

3. The root diameter of a f " U.S. Std. threaded screw is 
•6201". What is the pitch? 

4. The root diameter of a U.S. Std. threaded bolt is 3-567". 
If the pitch is |, what is the outside diameter of the screw? 

5. If the single depth of a U.S. Std. thread is -0491", find 
the pitch? 

6. If the double depth of a U.S. Std. thread is -3248", 
what is the number of threads per inch? 

7. The single depth of a U.S. Std. thread is -1998", what is 
the number of threads per inch? 

8. What sized tap drill would be used for a If" screw, U.S. 
Std. thread, 7 threads to the inch? 

9. What sized tap drill would be used for a If" screw, 
Std. thread, 5 threads to the inch? 

10. What sized tap drill would be used for a 1" screw, 
Std. thread, 8 threads to the inch? 

(3) Square Thread. 



U.S. 



U.S. 





Fig. 83 

The square thread is used in screws which are subjected to 
heavy loads, the jack-screw being an example. 

In this thread the sides are parallel, the thickness of the 
tooth, the depth, and the width of the groove being all theo- 
retically equal. In practice, however, the width of the groove 
is made slightly larger than the thickness of the thread to allow 
for clearance. 

The pitch — or the distance from the middle point of one 
tooth to the middle point of the next — is in the square thread 
equivalent to one tooth and one space. 



194 



MATHEMATICS FOR TECHNICAL SCHOOLS 



If, as in previous cases, we take a pitch of 1", then the thick- 
ness of the thread will be §", the depth \*\ and the width of the 
groove \". 

Example : 

Find the root diameter of a square thread 3" in diameter, 
with a pitch of \. If the pitch is |, then the depth is \" . 
.'. double depth = |". .\ root diameter = 2f". 

We could obtain the same result by using a formula similar 
to that for preceding threads. 

1 



Root Diameter = Outside Diameter — 
.'. Root Diameter = 3 " -\" = 2f*. 
(4) Acme 29° thread. 



No. of threads per inch 






c 


P' 5 ^ 


■•—pitch - 






♦ / \ 

* 1 \ 
a 1 \ 


P AM N r 
/? \ / 




\ g 




i H 


§ r V 








/ z 

Y T 


5 ' \ 








* ' 


B 







Fig. 84 

This thread was designed to overcome the defects in the 
square thread. It is less difficult to make and does away with 
the sharp corners. Its principle use is in machine tool manu- 
facture, where it is used for lead screws and other service 
where power is transmitted. 

The angle between the threads is 29°, and theoretically the 
depth of the thread is one half the pitch. 

If we consider the figure to the right above — pitch 1" — 
we have : 

In A ABC, BC= -5" Cot 14° 30' 
= -5"X3-86671 
= 1-93335" 
.'. 2 43f = l-93335"--5" = l-43335" 
.*. AM= •71667". 



MATHEMATICS OF THE MACHINE SHOP 195 

In A ADM, DM = AM tan 14° 30' 

= • 71667 "X -25862 
= • 185345 " 
.-. 2 DM or DN = -37069" 
.-. Width of flat at top = -3707 " 
Width of flat at bottom = -3707" 
Width of space at top = l"- -3707"= -6293". 

These are constants for all pitches. In practice to give 
clearance the following measurements are used (P = pitch): 

Width of flat at top = -3707 P 
Width of flat at bottom = -3707 P- 0052" 
Width of space at top = • 6293 P 

Depth of thread = \ P + • 010". 

Exercises LXXVII. 

1. Find the outside diameter of a screw with a square thread 
which has a root diameter of 1" and a pitch of ^. 

2. Find the root diameter of a square thread which has an 
outside diameter of 2\" and a pitch of \. 

3. A square thread has an outside diameter of 4" and an 
inside diameter of 3|". What is the pitch? 

4. Find the root diameter of a square thread which has an 
outside diameter of 3|" and a pitch of \. 

5. By a method similar to that employed in the preceding 
for a 1" pitch, find the width of flat at top, width at bottom, 
width of space at top, when the Acme 29° thread has a pitch 
of \. Check by means of data furnished for 1" pitch. 

6. If the depth of an Acme 29° thread is -3850", what is the 
number of threads per inch? 

7. If the width of flat at the top of an Acme 29° thread is 
• 1853", what is the pitch? 

8. If the width of space at the top of an Acme 29° thread is 
•1573", what is the number of threads per inch? 

9. If the width of space at the bottom of an Acme 29° is 
•0566", what is the number of threads per inch? 



196 



MATHEMATICS FOR TECHNICAL SCHOOLS 



(5) Whitworth Thread. 




1 \ < 


1 

it 


11 1 Q 






u 














11/ - 


8 




Fig. 85 

This is a standard thread in England and on the Continent. 

The sides form an angle of 55° with one another, while the 
top and bottom are rounded. The rounded part at both top 
and bottom is equal to one-sixth of the total depth of triangle 
above, leaving two-thirds for the depth of the thread. 

In above figure if AD = x, BC = radius of rounded part (r), 

then AB=?+r. 
o 

If the pitch be 1" we have: 

In &ADE, AD= -5 Cot 27° 30' 

= -5"X 1-92098 
= • 96049 " = x. 

x i 
6 +r ' 



In A ABC, Cosec 27° 30' 



2-16568 = 



6 +r 



2-16568 r = -+r 
6 

1-16568 r = f = 
6 



96049 
6 



= •16008" 



.'. r=-1373" 

depth of thread = f X • 96049" = • 64033". 

The dimensions of this thread stated in terms of the pitch 
(P) are as a consequence of the above: 

Depth = -64033 P. 

Radius of rounded part = • 1373 P. 

Example: 

Find the depth and radius of curvature of a Whitworth 
thread having a pitch of T V 

Depth = -64033 X T V = -064033". 

Radius of Curvature = • 1373 X T V= ■ 01373 *. 



MATHEMATICS OF THE MACHINE SHOP 



197 



Exercises LXXVIII. 

1. Find the depth and radius of curvature of a Whitworth 
thread having a pitch of yj. 

2. The depth of a Whitworth thread on a f " screw is • 064033 " 
"What is the pitch and the diameter at the root? 

3. A 1" screw has a Whitworth thread with a pitch of • 1250". 
What is the depth of the thread and the diameter at the root? 

4. A 1\" screw with a Whitworth thread has a diameter at 
the root of 1-067". Find depth of thread, the pitch, and radius 
of curvature. 

5. The depth of a Whitworth thread on a 2" screw is • 1423". 
Find the number of threads per inch, the diameter at the root, 
and the radius of curvature. 

6. A |" screw has a Whitworth thread with 16 threads per 
inch. Find the depth of the thread, the diameter at the root, 
and the radius of curvature. 

129. Thread Cutting. 

Gear Trains. One of the common ways of transmitting 
motion from one point to another is by means of gear trains. 




Fig. 86 



The simplest form of gear train, having but two gears, is 
shown in Figure 86. Gears are usually known by their number 
of teeth. Thus, if / has 20 teeth it would be called a 20-toothed 
gear. Similarly // would be called a 60-toothed gear. 



198 



MATHEMATICS FOR TECHNICAL SCHOOLS 



If two such gears are in mesh, as above, and the motion 
from 7 is transmitted to 77, 7 would be known as the driver 
and 77 as the driven. As each tooth in 7 pushes along a corres- 
ponding tooth in 77, it follows that one revolution of 7 will 
cause 77 to make only one-third of a revolution. Therefore the 
shaft to which 7 is keyed will make three revolutions while the 
shaft to which 77 is keyed is making one revolution. This 
principle is used extensively in gear trains. 




Fig. 87 

In Figure 87 we have three gears in the train. It may be 
necessary to insert the intermediate gear 77, either, that 7 
and 777 may have the same direction, or to permit of 7 driving 
777 without increasing the size of the gears. The gear 77 has 
no effect on the speed ratio of 7 and 777, for when 7 moves one 
tooth the same amount of motion will be transmitted to 77, 
which in turn will move 777 one tooth. Since each revolution of 
7 will result in one-fourth of a revolution of 777, therefore the 
speed ratio of 7 to 777 will be 4 to 1. 

This may be stated as follows: 

R.P.M. of driver _ teeth on driven 80 _ 4 
R. P.M. of driven teeth on driver 20 1" 

Frequently it is necessary to make such a great increase or 
decrease in speed, that to accomplish it with a simple train of 



MATHEMATICS OF THE MACHINE SHOP 



199 



gears, would necessitate too great a difference in diameters. 
For this purpose a compound gear train is used. 




Fig. 88 

Figure 88 represents a common form of a compound gear 
train, I drives II and causes a reduction of speed, /// is keyed 
to the same shaft as //and therefore travels at the same speed, 
/// meshes with IV, and on account of their relative number 
of teeth, a further reduction of speed is effected. 

If we wish to find the speed ratio of / and IV we might 
proceed as follows: 

Since / has 20 teeth and // 40 teeth, the speed of / is twice 
that of //. Since /// has 20 teeth and / V 80 teeth, the speed 
of ///, that is of //, is four times that of IV. Combining 
these statements we have that the speed of / is eight times that 
of IV. 

The above is equivalent to the following: 

R.P.M. of first driver Product of No. of teeth of all the driven 



R.P.M. of last driven Product of No. of teeth of all the drivers 

In above figure / and /// are the drivers and // and IV 
the driven. 

Substituting the values from the figure in the above relation: 

R.P.M. of first driver 40X80 8 
R.P.M. of last driven ~ 20X20 " 1* 



200 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Cutting a Thread. If a piece of work, on which a thread is 
to be cut, is placed in a lathe, it will revolve at the same rate 
as the spindle. If the spindle and lead screw turn at the same 
rate, then the number of threads per inch on the work will 
be the same as the number of threads per inch on the lead 
screw. If it is necessary that the number of threads per inch 
on the work differ from the number of threads per inch on the 
lead screw, then the principle of changing the speed by in- 
serting gears of different sizes becomes necessary. 



SPINDLE. GEAR 



CHANGE STUD GE AP. 




INSIDE STUD GEAP. 



■• MlliWiW tW till 



X LEAD SCREW 
^ \CHANQE LEAP 5CBEW GEAR. 



Fig. 89 

Figure 89 shows the relation of gears in a simple geared 
lathe. 

The spindle gear turns with the spindle, and drives the inside 
stud gear through the idler. The change stud gear, which is 
keyed to the inside stud, transfers the motion through another 
idler to the lead screw. 



MATHEMATICS OF THE MACHINE SHOP 201 

If the spindle gear in the above has 24 teeth, the inside gear 
on the stud 24 teeth, the change stud gear 40 teeth, and the 
lead screw 80 teeth, then: — Speed of spindle ff of speed of 
stud. Speed of stud f§ of speed of lead screw. .*. Speed of 
spindle = ftXf§ = f of speed of lead screw. 

The same result may be obtained by substituting in the 
formula, giving: 

Revolutions of spindle _ Product of No. of teeth in driven 
Revolutions of lead screw Product of No. of teeth in drivers 

= 24X80 2 
" 24X40" 1' 

In this case if the lead screw has 6 threads per inch, then the 
work would have 12 threads per inch. 

Knowing the Lead of the Lathe we can find an arrangement 
of gears which will give the desired number of threads per 
inch on the work. 

Example: 

If the lead of the lathe is 8, find the necessary gears on stud 
and lead screw to cut a thread of T V pitch. 

In this case the lead screw will advance |" in one 
revolution and we want the work to advance ^ in one 
revolution. 

This ratio of 8 to 10 would be obtained if we placed an 
8-toothed gear on the stud and a 10-toothed gear on the lead 
screw. These gears are, however, not obtainable, but the same 
ratio may be maintained if we place a 48-toothed gear on the 
stud and a 60-toothed gear on the lead screw. 

Gears furnished with a Lathe. Gears for a lathe usually 
vary in size by adding the same number of teeth each time 
to the gear just below. The two common sets are those 
obtained by adding 4 to the one below, giving 24, 28, 32 ... . 120, 
and those obtained by adding 7, giving 21, 28, 35.... 105. 
This is called gear progression. 



202 MATHEMATICS FOR TECHNICAL SCHOOLS 

Exercises LXXIX. 

1. A lead screw has 6 threads per inch. What gears must 
be placed on stud and lead screw to cut 16 threads per inch? 

2. Determine the change gears for cutting a ^ pitch thread 
when the lead screw has a § pitch. 

3. A lathe with a lead screw of £ pitch has a 24-toothed gear 
on the stud and a 60-toothed gear on the lead screw. How 
many threads will be cut on a screw when the carriage has 
advanced 3 inches? 

4. How many threads per inch will be cut by a lathe when 
the lead screw has a 64-toothed gear and the stud a 24-toothed 
gear, the lead screw having a £ pitch? 

5. We wish to cut 24 threads per inch on a lathe with a 
lead of 8 and a gear progression of 4. What gears would 
be used? 

6. The lead screw is | pitch, the screw to be cut j? pitch. 
If there is a 24-toothed gear on the stud, what gear must be 
placed on the lead screw? 

7. A lathe having a 72-toothed gear on the lead screw and a 
24-toothed gear on the stud cuts 18 threads per inch. What is 
the pitch of the lead screw? 

8. The lead screw has a 96-toothed gear and a £ pitch. What 
gear on the stud will cut 24 threads per inch? 

9. What gear must be used on the lead screw in order to 
cut 12 threads per inch, when the lead screw has 6 threads per 
inch and a 36-toothed gear is used on the stud? 

10. We wish to cut 14 threads per inch on a lathe with 
a lead of 6 and a gear progression of 7. What gears would 
be used? 

Compound Gearing in the Lathe. Owing to the limited 
number of gears and also to give a wider range of speeds to 
those available, the principle of compound gears is used on the 
lathe. 

Figure 90 represents the arrangement of gears on a lathe 
when compounding is necessary. 



MATHEMATICS OF THE MACHINE SHOP 



203 



The only difference between this arrangement and that 
described in the simple geared lathe is that instead of the idler 
which meshes with the lead screw gear, there are two gears 
keyed to the same shaft. The inside one has usually twice as 
many teeth as the outside, and as a consequence causes an 



SPINDLE GEAE. 



1=^2- 



CHANGE STUPGEAg , 



OUTSIDE COM- 
POUNDtNQ GEAB 




INSIDE STUD GEAe 



■E 



wii i iiwtw mwtffl* - 



\ LEAD SCREW 



jj frChANGE LEAP 5CREW GEAg. 

Fig. 90 



additional reduction of speed in the ratio of 2 to 1. By the use 
of this compound the same gears which are used to cut 9, 
10, 12, etc., threads on the simple geared lathe will cut 18, 20, 
24, etc., threads. 

Example 1: 

We wish to cut 24 threads per inch on a lathe with a lead of 6. 

We may use the same gears as in the example under the 
simple geared lathe, i.e., a 24 on both spindle and inside stud, 
a 40 on the change stud, and an 80 on the lead screw gear. If 
now we place a compound between the change stud and lead 



204 MATHEMATICS FOR TECHNICAL SCHOOLS 

screw gear, consisting of a 72 and a 36 (see diagram), we would 
have the following speed ratio: 

Revolutions of spindle _ Product of No. of teeth in driven 
Revolutions of lead screw Product of No. of teeth in drivers * 

24X72X80 4 
= 24X40X36 ~ 1' 

In this arrangement the spindle will make four revolutions 
when the lead screw is making one, therefore 24 revolutions 
when the lead screw is making 6 revolutions. 

Example 2: 

We wish to cut 3 threads per inch on a lathe with a lead of 6. 
In this case it is necessary for the spindle to revolve only one- 
half as fast as the lead screw. For this purpose we might use 
the simple gear with say an 80 on the change stud and a 40 on 
the lead screw. We might also use the compound gear with 
equal gears on the change stud and lead screw, say 40 and 40, 
and interchange the gears on the compound, i.e., have the 
change stud mesh with the small gear on the compound and 
the lead screw mesh with the large gear on the compound. 
Then as in preceding cases: 

Revolutions of spindle _ Product of No. of teeth in driven 
Revolutions of lead screw ~~ Product of No. of teeth in drivers ' 

_ 24X36X40 _ 1 = 3 
"24X72X40 2 6 - 

In practice the machinist reduces the method of finding the 
necessary gears when compounding to the following rule: 

''Write the ratio of the speed of the driving gear to the driven 
gear as a fraction, divide the numerator and denominator into 
two factors and multiply each pair of factors by the same number 
until gears with suitable number of teeth are found. The gears 
in the numerator are the driven and those in the denominator 
the driving gears." 






MATHEMATICS OF THE MACHINE SHOP 205 

Applying this rule to the two examples above, we have 
In Example 1: 

24 6X4 /6X12 X _ /4X20n 



6 3X2 



/ DX1A / 3:XZU \ 

~ V3X12/ X V2X20;' 



= 72 80 = 4 
36 40 1' 



In Example 2: 

3 3X1 /3Xl2x /1X40 



6 6X 



1 _ / dXIA / lAt» \ 

1 ~ V6X12/ X V1X40J 



= 36 40 = 1 
72 40 2 



Reduction Gears in the Head-stock. Some lathes, par- 
ticularly those intended for cutting fine threads, have reduction 
gears in the head-stock. If in this case equal gears are placed 
on the change stud and lead screw, the spindle does not make 
the same number of revolutions as the lead screw. The ratio 
of this gearing in the head-stock is usually 2 to 1, so that with 
equal gears on the change stud and lead screw the spindle will 
turn twice as fast as the lead screw. In such lathes this must 
be taken into account in figuring the necessary gears. 

Cutting of Double, Triple, etc., Threads. To cut a double 
thread on a screw, say 8 per inch, we would set the lathe for 
cutting half that number, in this case 4. Having cut this, turn 
the work one-half of a revolution and repeat the operation. 

To cut a triple thread, set the lathe for cutting one-third the 
number. Having cut this, turn the work one-third of a 
revolution and repeat. 

Exercises LXXX. 

1. A lathe has a lead screw with a lead of 4, and has a 40- 
toothed gear on the stud and a 90-toothed gear on the lead 
screw. Using a 72 and 36 as compounding gears, "how many 
threads are cut? 

2. We wish to cut 11| threads per inch on a lathe w r ith a 
lead of 6. If the gear progression is 4, what gears would do 
the work by compounding? 



206 MATHEMATICS FOR TECHNICAL SCHOOLS 

3. We wish to cut If threads per inch on a lathe, the lead 
screw having 6 threads per inch. If the gear progression is 4, 
what gears would do the work by compounding? 

4. We wish to cut 64 threads per inch on a lathe with a lead 
screw having 8 threads per inch. If a 24-toothed gear is used 
on the stud, what gears placed on compound and lead screw 
would do the work? 

5. A lathe has a lead of 6. If the gear progression be 7, 
calculate the change gears for cutting 14 threads per inch. 

6. What gears must be used to cut 12 threads per inch on a 
lathe having a lead of 6, when 36 and 72 are used as com- 
pounding gears? 

7. A special job requires 2\ threads per inch. If the lathe 
has a lead of 4, what gears would do the work? 

8. In a boat-lifting apparatus a 12-toothed gear meshes with 
a 48-toothed gear. Keyed to the latter is a 12-toothed gear, 
which meshes in turn with another 48-toothed gear on the 
revolving shaft. If the revolving shaft is 3 in. in diameter, 
how many turns of the handle will be necessary tofraise the 
boat h\ feet? 

9. A lathe has 6 threads per inch on the lead screw and a 40 
and 80 on the inside and outside compound respectively. 
What gears must be used on stud and lead screw to cut 3 
threads per inch? 

10. It is desired to cut 4 threads per inch on a piece of work 
The lead screw gear has 6 threads per inch, while a 36-toothed 
gear is placed on the stud and a 48-toothed gear on the 
lead screw. What arrangement of compound gears would be 
suitable? 

Quick Change Gears. To avoid the difficulty of having to 
calculate the necessary change gears, modern lathes are 
equipped with a mechanism for this purpose. 

In Figures 91 and 92 this mechanism is shown. 

The device is complete in one unit, and is contained in a 
box which is mounted on the front of the bed where its operating 
levers are convenient to the operator. The mechanism consists 
essentially of a cone of gears, an intermediate shaft, and a set 
of sliding gears. The tumbler gear is permanently in mesh with 
a long face pinion located inside the barrel about which the 



MATHEMATICS OF THE MACHINE SHOP 207 

tumbler gear pivots. This gear may be tumbled into engage- 
ment with any of the nine gears in the cone, thus imparting 




Fig. SI 




Fig. 92 

nine changes of speed to the intermediate shaft which is 
permanently geared to the cone. It will thus be seen that 
thirty-six changes are obtained with two operating levers and 
without removing any of the gears. 

130. Gear Calculation. In the last section the principle of 
change gears as applied to the lathe was dealt with. We will 
now consider some calculations pertaining to the gear itself. 



208 



MATHEMATICS FOR TECHNICAL SCHOOLS 



In Figure 93 some of the more important terms with respect 
to a spur gear are indicated. 



CIRCULAR. PITCH 
TH I CKNE2>5 OF TOOTH 



PITCH CIRCLE 



ADDENDUM 

DEDENDUM 

WOKWNQ DEPTH 
-WHOLE DEPTH 




h— ROOT DIAMETER-*- 

' PITCH DIAMETER.— 

- OUTolDB DIAMETER. 

Fig. 93 

The Pitch Circle is the line half-way between the top and 
bottom of the teeth. When two spur gears mesh, their pitch 
circles are regarded as being in contact. 

The Pitch Diameter is the diameter of the pitch circle. 

The Diametral Pitch is the number of teeth to every inch of 
pitch diameter of the gear. If the gear has 36 teeth and is 
4 in. in diameter, it is said to be a 9 pitch gear. 

The Circular Pitch is the distance from the centre of one 
tooth to the centre of the next, measured along the pitch circle. 

The Thickness of the tooth should be slightly less than the 
space between the teeth to allow for clearance, but in practice 
they are calculated as being equal. As a result either the 
tooth or the space is one-half the circular pitch. 



MATHEMATICS OF THE MACHINE SHOP 209 

Clearance must be provided at the bottom of the space 
between the teeth (see diagram). It is usually ^ of the thick- 
ness of the tooth measured on the pitch circle. 

The Addendum is the part of the tooth projecting beyond 
the pitch circle. It is reckoned as a fraction of the size of the 
tooth. Thus in a 12 pitch gear the addendum would be fa*. 

The Dedendum is the part of the tooth between the pitch 
circle and the working depth. 

The addendum plus the dedendum make the working 
depth of the tooth. 

The Root Diameter is the diameter measured at the bottom 
of the space (see diagram). 

The Outside Diameter is the diameter measured at the 
outside of the gear (see diagram). 

Knowing the number of teeth in a gear and the diametral 
pitch, to find the size of gear blank, i.e., outside diameter. 

Example : 

What should be the outside diameter of a gear blank for a 
gear of 98 teeth and a diametral pitch of 14? 

Diameter of pitch circle = ff = 7 ". 

Addendum = 3^" on one side. 

= \" on both sides. 

. • . Outside diameter = 7 " + \ " = 7 • 1428 ". 

To Find the Depth of Cut necessary in the Preceding Example. 

Total depth = Addendum -f- Dedendum + Clearance. 

Since the clearance depends on the thickness of the tooth 
it will first be necessary to determine the thickness. 

Number of teeth = 98. 

Since there are 14 teeth for 1" of diameter there will be 14 
teeth for 3-1416" of circumference. 

.'. Circular pitch = 3 ' |^ 16 = -2244". 
14 



210 MATHEMATICS FOR TECHNICAL SCHOOLS 

Since the circular pitch is the distance which a space and 
tooth together occupy, 

.'. thickness =| of -2244" = 1122 ". 

Since clearance = -^ of thickness, 

.'. clearance in above = -01122". 

.'. Total depth = ^"+^"+ -01122"= • 15407". 

Exercises LXXXI. 

1. The circular pitch of a gear is -3927". What is the 
diametral pitch? 

2. The diametral pitch is 12. Find the circular pitch. 

3. Find the thickness of tooth on a 14 pitch gear. 

4. Find the total depth of tooth on a 14 pitch gear. 

5. Find the thickness of tooth on a 16 pitch gear. 

6. Find the total depth of tooth on a 16 pitch gear. 

7. Find the outside diameter of a gear blank for a 60-toothed 
gear, 12 pitch. 

8. Find the outside diameter of a gear blank for 20 teeth 
with a circular pitch of -7854". 

9. What is the number of teeth on a gear 6" outside dia- 
meter, 12 pitch? 

10. What is the number of teeth on a gear 8" outside diameter, 
6 pitch? 

11. What is the pitch of a gear having 63 teeth and measuring 
6-5" outside diameter? 

12. What is the distance between the centres of a pair of 
gears having 72 teeth and 54 teeth respectively, 9 pitch? 

131. The Milling Machine. "Milling is the process of 
removing metal with rotary cutters. It is used extensively in 
machine shops to-day for forming parts of machinery, tools, 
etc., to required dimensions and shapes. A machine designed 
especially for this purpose was in existence as early as 1818, 
but little progress was made in the process until after the 
invention of the universal milling machine in 1861-62 by 
Joseph R. Brown of J. 11. Brown and Sharpe." 



MATHEMATICS OP THE MACHINE SHOP 211 

132. Cutting Speed. In determining the cutting speed of a 
lathe we multiplied the circumference of the work in feet by 
the number of revolutions which the work made per minute. 
In the milling machine the diameter of the milling cutter 
corresponds to the diameter of the work in the lathe. The 
cutting speed of the milling cutter is therefore obtained by 
multiplying the circumference of the cutter in feet by the number 
of revolutions which it makes per minute. 

Thus if a milling cutter 6" in diameter makes 60 revolutions 
per min. the cutting speed = Circumference of cutter in ft. 
X Revolutions per min. = ^- X^X ^ =94+ft. per min. 

133. Feed. The feed on a milling machine is usually 
reckoned in inches per min. As in the case of the lathe only a 
general rule can be given. "In roughing, slow speed and 
heavy feed using a coarse-pitch cutter. In finishing, fast 
speed and light feed using a fine-pitch cutter." In Figure 94 
following, a coarse-pitch cutter and a fine-pitch cutter are 
shown: 





Fig. 94 



R. H. Smith in "Advanced Machine Work" gives the fol- 
lowing table for speeds and feeds: 
Speeds. 



212 MATHEMATICS FOR TECHNICAL SCHOOLS 

With Carbon Steel Cutters. Cast iron — 40 ft. per min. 
Machine steel — 40 ft. per min. Annealed carbon steel — 30 
ft. per min. 'Brass or composition — 80 ft. per min. 

With High Speed Steel Cutters. Cast iron— 80 ft. per min. 
Machine steel — 80 ft. per min. Annealed carbon steel — 60 ft. 
per min. Brass or composition — 160 ft. per min. 
Feeds. 

Feeds for milling cutters are from -002" to -250" per cutter 
revolution, and depend on diameter of cutter, kind of material, 
width and depth of cut, size of work and whether light or 
heavy machine is used. 

In order to calculate the feed it is necessary to know the 
lead of the feed screw and the number of revolutions per 
minute at which it is turning. Thus, if the lead of the feed 
screw is \", and it is turning at the rate of 3 revolutions per 
min., then the feed = |"X3 = f " per min. 

Exercises LXXXII. 

1. A milling cutter 4" in diameter is turning at a rate of 
40 R.P.M. What is the cutting speed? 

2. A milling cutter 3^" in diameter is cutting at a speed of 
36f ft. per min. What is the R.P.M. ? 

3. A milling cutter turning at a rate of 56 R.P.M. has a 
cutting speed of 60 ft. per min. What is the diameter of the 
cutter? 

4. A milling cutter 6" in diameter is cutting at a speed of 
66 ft. per min. What is the R.P.M.? 

5. A milling cutter 2" in diameter is running at 58 R.P.M. 
What is the cutting speed? 

6. A milling cutter turning at a rate of 30 R.P.M. has a 
cutting speed of 40 ft. per min. What is the diameter of the 
cutter? 

7. The feed screw in a milling machine is single threaded and 
has a pitch of £. If it is turned at a rate of 6 R.P.M., what is 
the feed? 

8. The feed screw in a milling machine has a double thread 
with a pitch of \. If it is turned at a rate of 4 R.P.M., find 
the feed. 



MATHEMATICS OF THE MACHINE SHOP 



213 



9. The feed screw on a milling machine has a lead of \" . 
How many R.P.M. does it make if the feed is \\" per min.? 

10. The feed screw on a milling machine has a feed of \\" 
per min., and is being turned at 6 R.P.M. What is the lead 
of the screw? 

134. Indexing. One of the purposes of the milling machine 
is to cut slots or grooves in a circular piece of work at regular 
intervals. It is, therefore, necessary that it should have an 
attachment for dividing the circumference of the work into 
equal parts. This attachment is called the dividing head. 
The process of dividing the work into equal parts is called 
indexing. 

The metnods of indexing may be classified as — Rapid 
Indexing, Plain Indexing, Differential Indexing. 

Rapid Indexing permits of only a limited number of divisions 
of the circumference of the work, plain indexing extends the 
number of divisions, while differential indexing permits of a 
still wider range. 



WEX PLATE 



/^INDEX SPINDLE 




Fig. 95 



135. In Rapid Indexing the. index plate is fastened directly 
to the nose of the spindle as shown in Figure 95. This plate 
usually has 24 holes and is rotated by hand to any desired 
position, being held in place by a stop-pin. 



214 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Assume that in Figure 96 we have a round-headed 
bolt which is required to be milled so that the head becomes 





Fig. 96 

square. In this case it is evident that the work must be 
turned through \ of a revolution when one side of the work has 
been milled and we are ready to mill the next. We would 
therefore turn the index plate \ of a revolution, i.e., 6 holes. 
Using this kind of indexing we may obtain any number of divi- 
sions which will divide evenly into 24, as two, three, four, six, etc. 




-iP=H} 



Fi(i. 97 

136. Plain Indexing. In the figure above the index spindle 
is shown with a worm and worm-wheel mechanism, the worm 



MATHEMATICS OF THE MACHINE SHOP 



215 



being attached to the crank turned when indexing. The worm- 
wheel is keyed to the index spindle to which the work is 
attached. 

The principle of Plain Indexing may be seen from the 
diagram above. In the majority of index heads the worm is 
single threaded and the worm-wheel has 40 teeth. If, therefore, 
the index crank is turned one complete revolution, the worm 
will make one revolution, which moves the worm-wheel one 
tooth or ^ of its circumference. If, therefore, we want to turn 
the worm-wheel, and hence the spindle to which it is attached, 
one full revolution, we must turn the index crank 40 revolutions. 
If we want to turn the spindle £ of a revolution, we will turn 
the index crank 8 revolutions, and so on. 

If now we assume that it is required to cut seven flutes 
equally spaced in a reamer, we would first insert the stop-pin 




Fig. 98 



(see diagram) and then estimate the required number of 
revolutions of the index crank. In order to index for each 
flute the index crank must be turned 4j£ =5f revolutions. 



216 MATHEMATICS FOR TECHNICAL SCHOOLS 

To accurately fix this f of a revolution the index plate is made 
with a series of holes arranged in concentric circles. A sample 
plate is shown (Fig. 98). 

It is required to choose some circle where the total number 
of holes can be divided by 7. In the plate shown the outside 
row having 49 holes will suffice. 

We will, therefore, turn the index crank five complete revolu- 
tions, afterwards turn to the 35th hole in the outside circle of 
holes and insert crank-pin. 

Most milling machines are furnished with three index plates, 

each having six index circles. The following numbers of holes 

in the index circles of the three plates are used : 

15 16 17 18 19 20 

21 23 27 29 31 33 

37 39 41 43 47 49 

Exercises LXXXIII. 

1. By the rapid method show how you would index for 
milling a hexagonal head on a round bolt, if the index plate has 
24 holes. 

2. A piece of work is to have eight sides regularly spaced. 
How would you index by the rapid method, if the index plate 
has 24 holes? 

3. What diameter must a piece be to mill square lj" across 
the flats? 

4. What diameter must a piece be to mill hexagonal 1\" 
across the flats? 

5. Twelve flutes are to be milled in a tap. How would you 
index, using plain indexing, assuming that 40 turns of the 
index crank are required for one turn of the spindle? 

6. It is required to cut nine regularly spaced flutes in a 
reamer. How would you index, assuming a ratio of 40 to 1 
between index crank and spindle? 

7. Assuming a ratio of 40 to 1 between index crank and 
spindle, find the number of complete turns, the proper plate, 
and the number of holes for indexing 15 divisions. 

8. If the ratio between index crank and spindle be 60 to 1, 
what indexing would be used for 21 divisions? 



MATHEMATICS OF THE MACHINE SHOP 



217 



9. If the ratio between index crank and spindle be 40 to 1, 
what indexing would be used to cut 84 teeth in a spur gear? 

10. If the ratio between index crank and spindle be 40 to 1, 
what indexing would be used to cut 105 teeth in a spur gear? 

11. It is required to cut 85 teeth in a spur gear. How would 
you index assuming a ratio of 40 to 1 between index crank and 
spindle? 



A 



JP/HPLE 



ZL^ 









tt----_- 



\i 



I 



Unjr & 



i / 



— i— Ll 



-a 



•! — I— 



t_zj 



■¥ 



I 




Fig. 99 



] 



137. Differential Indexing. Assume that we require to 
cut 101 teeth in a spur gear. If we proceed as in plain indexing 
we would conclude that the index crank must turn T 4 ^- revo- 
lutions for milling each slot. As this fraction will not reduce 



218 



MATHEMATICS FOR TECHNICAL SCHOOLS 



to lower terms it would be necessary to have a plate 
with a circle containing 101 holes. As such a plate is not 
available for all machines a different mechanism is 
necessary. The method employed in such cases is called 
differential indexing. 

The diagrams 99 and 99a will help to explain the 
method. 




Fig. 99a 



The index crank and index plate are shown connected with 
the worm and worm-wheel as in plain indexing. On the outer 
end of the spindle the gear a is fastened. To an adjustable 
bracket are keyed the two gears b and c. The gear b meshes 
with a and the gear c with an idler i, which in turn meshes with 
the gear d. Keyed to the same shaft with d is a bevel gear e, 
which meshes with another bevel gear f. 



MATHEMATICS OF THE MACHINE SHOP 



219 



If the index plate in the accompanying figure be held 
stationary then for one complete turn of the spindle it will 




Fig. 100 

be necessary for the crank-pin to pass the point marked P 
40 times. Suppose, however, that the index plate is left free 
to rotate and gears in the ratio of 1 to 1 be placed on the 
spindle and on the worm, then when the spindle is 
making one revolution the index plate makes one revolu- 
tion. If the gears be so arranged that the index plate turns 
in the same direction as the spindle, it would only be necessary 
for the crank-pin to pass P 39 times in order to turn the spindle 
once. If however the index plate turns in the opposite direction 
to the spindle, it is evident that the crank-pin must pass P 41 
times in order to turn the spindle once. This principle of 
having the index plate connected with gearing so that the crank 
makes other than Jfi turns for one complete turn of the spindle 
is the special feature of differential indexing. 

We will now return to the difficulty of cutting 101 teeth in a 
spur gear. Since our object is to turn the spindle y^-y of a 
revolution, and 101 is not a multiple of any of the numbers on 



220 MATHEMATICS FOR TECHNICAL SCHOOLS 

the index plates, we will select a number on either side of 101 
which is a multiple of some of the numbers. It is readily seen 
that 20 is a multiple of 100 and also that ^X 100 = 40. If, 
therefore, we make 100 moves of 8 holes each on the 20 hole 
circle we will turn the worm 40 revolutions. 

If 101 such moves be made we would have ^X101 = 40f 
rev. of worm. This is f of a revolution too many, which may 
be offset by moving the index plate in the opposite direction to 
the spindle by suitable gears. Splitting this ratio into two 
parts we have f=fX|. 

Since we cannot multiply these fractions by any numbers 
which will give gears in stock, we write — 

2_2 X 3 

= (3 X 24) X (5 X 8) 

= 48 24 
~72 X 40* 

The 48 and 24 will be placed on the drivers, i.e., a and c, the 
40 and 70 on the driven, i.e., b and d. It will be necessary to 
place one idler in the train, as in diagram, in order that the 
index plate may turn in the opposite direction to the spindle. 

Example 2: 

Required to cut 83 teeth in a spur gear. 

Our object here is to turn the spindle ^ of a revolution. 
Since 83 is not a multiple of any of the numbers on the index 
plates, we will select a number on either side of 83 which is a 
multiple of some of the numbers. Thus we observe that 16 
is a multiple of 80 and also that ^X 80 = 40. If, therefore, we 
make 80 moves of 8 holes each on the 16 hole circle, we will 
turn the worm 40 revolutions. 

If 83 such moves be made we would have ^-X83=4l£ 
rev. of the worm. As this is 1^ revolutions too many, the 
index plate must move opposite to the spindle. 



MATHEMATICS OF THE MACHINE SHOP 221 

As gears in the ratio of 3 to 2 are in stock, it is not necessary 
to split the ratio f , but write f = ff . 

Here the compound would be removed and the gears on 
worm and spindle connected by means of two idlers. The 
48 gear would go on the driver, i.e., on a and the 32 gear on the 
driven, i.e., on d. 

Example 3 : 

Required to cut 137 teeth in a spur gear. 

By trying different combinations as in the preceding cases 
we find that -^X 140 = 40. 

If, therefore, we make 140 moves of 6 holes each on the 21 
hole circle, we will turn the worm 40 revolutions. 

If 137 such moves be made we would have -£ T X 137 = 39^- 
rev. of worm. 

This lacks ^f- of a revolution, which may be offset by moving 
the index plate in the same direction as the spindle by suitable 
gears. As in the preceding case it is not necessary to split 
the ratio but use gears in the ratio of 6 to 7, i.e., 24 and 28. 
The 24 will go on the driver, i.e., on a, and the 28 on the driven 
i.e., on d. One idler would be inserted to connect the worm and 
spindle. 

Note. — When a simple gearing is used the number of idlers 
depends on whether the index plate is to turn in the same or 
opposite direction to that of the spindle. If in the same 
direction one idler will be inserted, if in the opposite direction 
two idlers. 

To obviate the necessity of working out these gears, tables 
are available giving the necessary gears for all required divisions 
of the work. 

138. Cutting Spirals. If the gears that drive the shaft 
carrying the worm gear be connected with the feed screw, then 
as the table advances the spindle will rotate. This will pro- 
duce a spiral cut in the work, such as may be seen in a spiral 
reamer or a twist drill. 



222 MATHEMATICS FOR TECHNICAL SCHOOLS 

The gears for this purpose are shown in the following diagram: 



ZnoGEAfcOh 




GEAROKbCREW 



Fig. 101 



The four change gears are indicated in the above figure. 
The screw gear and the first stud gear are the drivers, the others 



MATHEMATICS OF THE MACHINE SHOP 223 

being the driven. By using different combinations of change 
gears the ratio of the lengthwise motion of the table to the 
rotary motion of the spindle can be varied. 

139. Lead of the Machine. If the feed screw of the table 
has 4 threads to the inch, and 40 revolutions of the worm are 
necessary for one revolution of the spindle, then, if change 
gears of equal diameters are used, the work will make one 
revolution while the table advances 10 in. The lead of the 
machine is therefore 10 in. 

Some machines have feed screws with other than 4 threads 
to the inch, but the same principle as the above will give the 
lead of the machine. 

140. Calculating Change Gears. In calculating gears for 
screw cutting we had the following proportion: 

Product of teeth in driven _ Revolutions of spindle 
Product of teeth in drivers Revolutions of feed screw 

In a similar way we will now have: 

Product of teeth in driven _ Lead of spiral 
Product of teeth in drivers Lead of feed screw 

Example 1: 

Find the change gears for cutting a spiral with a lead of 20", 
when the lead of the machine is 10 ". 

tj Lead of spiral _ 20 _ 2 
' Lead of machine 10 1' 

Since four gears are used we split the ratio thus: 



1 1 A 1 Vl X 32> ,X Vl X 24>'~32 X 



24 
24 



We will, therefore, place 64 and 24 on the worm and 2nd 
stud respectively, and 24 and 32 on the 1st stud and feed screw 
respectively. 



224 MATHEMATICS FOR TECHNICAL SCHOOLS 

Example 2: 

Find the change gears for cutting a spiral with a lead of 
8-333", when the lead of the machine is 10". 
Lead of spiral _8^_5 
Lead of machine 10 6' 

5 5 1 

Splitting the ratio = = - X ^. 
o £ o 

_/ 5X20 \ / 1X24 \ 

V2X20/ X V3X247 

= 100 24 
~ 40 72' 
We will, therefore, place 100 and 24 on the worm and 2nd 
stud respectively, and 72 and 40 on the feed screw and 1st 
stud respectively. 

141. Position of Table in Cutting Spirals. In order that the 
cutter may have clearance in cutting the groove, it is necessary 
that the table of the machine should be set at an angle. This 
angle depends on two things: — The lead of the spiral and the 
diameter of the work to be milled. This angle may be deter- 
mined either graphically or by calculation. 




Fig. 102 



In the figure ABC is a right-angled triangle in which BC is 
equal to the lead of the spiral and AC the circumference of the 
work. The angle ABC will then be the required angle. 



MATHEMATICS OF THE MACHINE SHOP 225 

Finding the angle by calculation is however a more accurate 

method, thus: 

. . Circumference of work 
Tangent of required angle = Lead 

From trigonometrical tables this angle can readily be found. 
Example : 

Find the angle at which the table must be set in milling a 
twist drill 1" in diameter, lead 8-68". 

3- 1416X1 
If 6 be the required angle, then tan 6 = — ' — = -36193 

.-. = 19° 54' = 20° (approx.) 
Tables are available giving the proper gears and angle of 
the table for all necessary cases. 

Exercises LXXXIV. 

1. If the ratio between the worm and spindle of a dividing 
head is 40 to 1, find the differential indexing for the following 
divisions:— 83, 99, 111, 139, 159, 161, 171, 238, 269, 351. 
Verify from table. 

2. What is the lead of a milling machine if the feed screw 
has a lead of \" and the ratio of worm to spindle is 60 to 1? 

3. If the lead of a milling machine is 10", calculate the change 
gears for cutting spirals with the following leads: — 9", 1-067", 
1-200", 1-667", 2-200", 3-056", 4-000", 5-500", 6-482", 8". 
Verify from table. 

4. The following change gears were used in cutting a 
spiral, on worm 72, on 1st stud 24, on 2nd stud 24, on screw 
gear 48. If the lead of the machine was 10", what was the 
lead of the spiral? 

5. The following change gears were used in cutting a 
spiral, on worm 64, on 1st stud 24, on 2nd stud 32, on screw 
gear 40. If the lead of the machine was 10", what was the 
lead of the spiral? 

6. The following change gears were used in cutting a 
spiral, on worm 86, on 1st stud 24, on 2nd stud 24, on screw 
gear 40. If the lead of the machine was 10", what was the 
lead of the spiral? 

7. A spiral with a lead of 7-92" is to be cut on a gear blank 
with a pitch diameter of 3"; find the angle for setting the table. 



226 MATHEMATICS FOR TECHNICAL SCHOOLS 

8. A spiral with a lead of 9-34" is to be cut on a twist drill 
with a diameter of \\"; find the angle for setting the table. 

9. In milling a twist drill the table is set at an angle of 
15°, and the lead of the spiral is 11 -724 ". Find the diameter 
of the drill. 

10. In milling a twist drill the table is set at an angle of 
17° 30', and the diameter of the drill is If*. Find the lead 
of the spiral. 

Review Exercises LXXXV 

1. What is the lead on a double-threaded screw of \ pitch? 

2. A screw with a triple thread has a lead of 1". What is 
the pitch? 

3. How many revolutions must be made with a double- 
threaded screw, with a pitch of -j^, so that it may advance 2"? 

4. A sharp "V" thread with a pitch of ^, makes 6 turns to 
the inch. How is it threaded? What is the double depth 
of the thread? 

5. If the double depth of a sharp "V" thread is -1924", 
what is the number of threads per inch? 

6. A If" bolt with a sharp "V" thread has a diameter at 
the root of 1 • 2784". What is the depth of the thread? What 
is the pitch? 

7. How long will be necessary to take a cut over a shaft 
3' long and 2" in diameter with a feed of 18 and a speed of 36 
ft. per min? 

8. A drill cuts f " into a piece of work in 10 min. If it 
makes 40 revolutions per min., what is the feed of the drill? 

9. A piece of work 6" long is to have end diameters of 
•7432" and -6182"; find the amount of taper and also the angle 
of taper. 

10. A screw with a U.S. Std. thread has 16 threads to the 
inch. It has a diameter at the root of -2936". What is the 
diameter of the screw? 

11. A 6" screw with a Whitworth thread has a pitch of f. 
What is the root diameter of the thread? 

12. The width of the flat at the top of an Acme 29° thread 
is -0371". What is the pitch? 

13. A lathe has an 84-toothed gear on the lead screw and 
the pitch of the lead screw is £. What gear on the stud will 
cut 18 threads per inch, simple gearing? 



MATHEMATICS OF THE MACHINE SHOP 227 

14. A simple geared lathe having a 96-toothed gear on the 
lead screw and a 32 on the change stud cuts 24 threads per 
inch. What is the pitch of the lead screw? 

15. The lead screw on a lathe is j pitch, the screw to be 
cut -^ pitch, and the change stud gear has 24 teeth. If the lathe 
be simple geared what gear must be placed on the lead screw? 

16. If the lathe in the preceding question had reduction 
gears in the head-stock in the ratio of 1 to 2, what gear would 
be necessary on the lead screw? 

17. In a lathe with a lead screw of ^ pitch a 40-toothed 
gear is placed on the change stud and a 75 on the lead screw. 
If the lathe be simple geared how many threads per inch will 
be cut on the screw? 

18. How many threads per inch will be cut when the lead 
screw gear is 100, the change stud 60, the outside compound 
24, the inside compound 48, and the pitch of the lead 
screw |? 

19. A lathe with a 60-toothed gear on the change stud and 
a 40 on the lead screw, has 6 threads per inch on the lead screw. 
What compound gears are used in cutting 2 threads per inch? 

20. What should be the outside diameter of a gear blank 
for a gear of 24 teeth and a pitch diameter of 4? 

< 21. Two gears in mesh have 66, teeth and 88 teeth respec- 
tively. If they are 14 pitch gears, what is the distance between 
their centres? 

22. A milling cutter turning at the rate of 40 R.P.M. has 
a cutting speed of 40 ft. per min. What is the diameter of 
the cutter? 

23. What is the lead of a milling machine if the feed screw 
has a lead of £ and the ratio of worm to spindle is 40 to 1 ? 

24. Find the change gears for cutting a spiral with a lead 
of 1-64", when the lead of the machine is 10". 

25. A spiral with a lead of 7- 5 "is to be cut on a twist drill 
with a diameter of \\"\ find the angle for setting the table. 



CHAPTER XV. 

LOGARITHMS. 

142. If we wish to multiply 100 by 1000 we may do so in 
either of two ways : 

(1) 100X1000=100000 

(2) 100 = 10 2 and 1000 = 10 3 

/. 100 X 1000 =10 2 X10 3 = 10 5 = 100000. 

In the second method we observe that the product is obtained 
by adding the exponents of the powers of 10 which equal 100 
and 1000. 

If then we had numbers expressed as powers of 10, it would 
be possible to multiply them together by adding their 
exponents. 

Thus, if we wished to multiply 23 by 432 we might do so 
by addition of the exponents of the powers of 10 which equal 
23 and 432. 

We are here met by two difficulties. 

(1) What powers of 10 equal 23 and 432 ? 

(2) What number is represented by 10 when raised to 
the sum of these two powers ? 

Let us consider the following set of numbers: 

(1) 10, (2) 25, (3) 100, (4) 365, (5) 1000, (6) 7628, (7) 10000. 

We know that in (1) 10 = 10!, and that in (3) 100 = 10 2 . 

Now in (2) 25 is greater than 10, or 10 1 , and loss than 100, 
or 10 2 , therefore 25 = 10 1+adec,mal - 

Again in (4) 365 is greater than 100, or 10 2 , and less than 
1000, or 10 3 , therefore 365 = 10 2+adeclmal - 

Further in (6) 7628 is greater than 1000, or 10 3 , and less 
than 10000, or 10 4 , therefore 7628 = 10 3+adeclmal - 

228 



LOGARITHMS 229 

Tables have been worked out giving the decimal parts of 
the powers of 10 in the above. 

Thus, from the tables 25 = 10 1 3979i - 

365 = 10 256229 - 

7628 = 10 3 - 88241 - 

This exponent of the power to which we must raise 10 to 
give the number is called the logarithm of the number. 

Thus, logarithm of 25 = 1-39794. 
logarithm of 365 = 2-56229. 
logarithm of 7628 = 3-88241. 

In this system — called the Briggs' System — the base is 
10, and all numbers are considered as powers of 10. 

The contraction "log" is used instead of logarithm. 

143. Characteristic and Mantissa. 

In25=10 139794 ' 
the 1 in the exponent is called the Characteristic and the 
•39794 the Mantissa. The Mantissa is always positive. 

Characteristic written at sight. 

In the above set of numbers we observe that 25 which 
is greater than 10 and less than 100, has 1 for its characteristic; 
that 365 which is greater than 100 and less than 1000 has 2 
for its characteristic; that 7628 which is greater than 1000 
and less than 10,000, has 3 for its characteristic. We, there- 
fore, infer that the characteristic of the logarithm of any number 
greater than 1 is one less than the number of integral figures 
in the number. 

144. How to find the Logarithm of a number from the 
Tables. 

Find the log of 36. 

As previously explained we at once write down the charac- 
teristic, 1. 



230 MATHEMATICS FOR TECHNICAL SCHOOLS 

To get the decimal part we go down the left-hand column 
to 36, then along the horizontal row to the right, and under 
the vertical column headed 0, we read -55630, 

.-.log 36 = 1-55630. 

Find the log of 365. 

The characteristic here is 2. 

To get the decimal part we go down the left-hand column 
to 36 as before, then along the horizontal row to the right, 
and under the vertical column headed 5, we read • 56229. 

.-.log 365 = 2-56229. 

Find the log of 3658. 

The characteristic here is 3. 

To get the decimal part we proceed as in the last case, 
giving 3-56229. To make the adjustment for the 8, we follow 
the same horizontal row out to the mean differences. In the 
vertical column headed 8, we read 95. This we add to 3 • 56229, 
giving the log of 3658 = 3-56229 

95 



3-56324 
Find the log of 36587. 
The characteristic here is 4. 

To get the decimal part we proceed as in the last example, 
giving 4 • 56324. To make the adjustment for the 7, we observe 
that in the same horizontal row, under 7 in mean differences, we 
have 83. Since the 7 is in the fifth place, it has only one-tenth the 
value that it would have in the fourth place, therefore we move 
the 83 one place to the right before adding, thus 4 • 56324 

83 



4-563323 
.*. log 36587=4-563323 = 4-56332 to 5 places. 



LOGARITHMS 231 

145. Position of Decimal Point. Since the division of a 
number by 10 or 100 is made by moving the decimal place to 
the left, the position of the decimal affects the characteristic 
only. 

Thus, log 3658 =3-56324. 
log 365 -8 =2-56324. 
log 36-58 =1-56324. 
log 3-658 = 0-56324. 

146. Knowing the Logarithm of a Number to find the 
Number. 

Tables, called antilogarithms, have been worked out which 
enable us to find a number if we know its logarithm. 

If log x = 2-34563, find x. 

Since only mantissas are recorded in the tables, the charac- 
teristic 2 has no bearing on what we look up, but only serves 
to fix the decimal place in the result. 

Proceeding with -34563 in antilogarithms, just as outlined 
in finding a logarithm, we have 22131 

31 
15 



221635 

Since the characteristic of the logarithm of the required 
number is 2, it must have three figures in the integral part, 

.'. x = 221 -635 = 221 -64 to 5 figures. 

Returning to the difficulty raised when we wished to multiply 
23 by 432 we have: 

23 = 10 136173 - 
432= 10 2 - 63548 - 
.-.23X432 = io 1 - 3617 ^ 2 - 63548 = lO 3 " 72 ^ 9936 -03 = 9936 to 5 
figures. 

In practice the base 10 is not written down, but only the 
exponents. 



232 MATHEMATICS FOR TECHNICAL SCHOOLS 

Example : 

Find the value of 45-236X31-341. 
log 45-236 = 1-65514 log 3 1-341 = 1-49554 

29 • 55 

57 14 



1-655487 1-496104 

Sum of logs = 1-655487 
1-496104 

3-151591=3-15159 to 5 places. 

Antilog 3-15159= 14158 

16 
30 



1417-70 
.-. 45 -236X31 -341 = 1417 -7 to 5 figures. 

Exercises LXXXVI. 

Employ logarithms to find the value of: 

7-53X20-08X14-93. 

146-32X78-49X10-09. 

9-36X4-592X3-61X1-08. 

8-99X61-3X7-6297X3-92 

5-037X236-84X1-009. 

147. Logarithms Applied to Division. 

We have learned by the foregoing that to multiply two 
numbers together, we add their logarithms and find the anti- 
logarithm of the result. 

Since division is the reverse of multiplication, we could 
without further detail infer that, to divide one number by 
another, we subtract their logarithms and find the antilogarithm 
of the result. 



1. 


53X82. 


6. 


2. 


10-64X150. 


7. 


3. 


483-26X108. 


8. 


4. 


381-56X17-928. 


9. 


5. 


493-75X4-73. 


10. 



LOGARITHMS 233 



Thus, divide 365 by 73. 
log of 365 = 2-56229 
log of 73 = 1-86332 



difference 


= -69897 


antilog of 


•69897= 


=49888 
103 
80 



4 • 99990 .-. 365 -r- 73 = 4 • 9999. 

Example : 

_ ... , .43-21X148-92 
Find the value of U9>7X37 . 42 

log 43 • 21 = 1 • 63548 log 148 • 92 = 2 • 17026 

10 265 

59 

1-63558 



2-172969 



Sum of logs of numbers in numerator = 3 • 808549 (a). 
log 149-7 = 2- 17319 log 37 -42 = 1-57287 

206 23 



2-17525 1-57310 

Sum of logs of numbers in denominator = 3 -74835 (6). 
(a) -(b) =3-808549 
3-74835 



•060199= -06020 to 5 places. 
Antilog -06020= 11482 



1-1487 
.*. result = 1 • 1487 to 5 figures. 



234 MATHEMATICS FOR TECHNICAL SCHOOLS 

An abbreviated arrangement of the work is as follows- 

log 43 • 21 = 1 • 63548 l og 149 - 7 = 2 - 173 19 

10 206 

log 148-92 = 2- 17026 log 37- 42 = 1-57287 

265 9 o 

59 :_ 

3^08^9" 3 * 74835 

subtract 3 • 74835 

0-060199 
anti 0-06020 = 11482 
5 

1-1487 

148. Logarithm of a Number Less than Unity. 
We have the following: 

• 01 = — = — = 10~ 2 
100 10 2 1U 



•001 = 



= V7^=10- 



1000 10 3 



• 0001 = io5oo = 1 l= 10 - 4 

By our definition of logarithms we have from the above 

log -1=-1. 

log -01 = -2. 

log .001 = -3. 

log -0001 = -4. 

Consider the following set of numbers : 

(D '*• (3) -01. (5) -001. (7) -0001. 

(2) -06. . (4) -008. (6) -0007. 

We know from the above that in (1) -l^lO" 1 and that in 
(3) • 01 = 10- 2 . 



LOGARITHMS 235 

Now in (2) -06 is greater than -01, or 10~ 2 , and less than 
•1 or 10" 1 , therefore -06 = i(r 2 + adeclmal - 

Again in (4) -008 is greater than -001, or 10~ 3 , and less 
than -01, or 10" 2 , therefore -008 = i()- 3 + adeclmaI - 

Further in (6) -0007 is greater than -0001, or 10 -4 and less 
than -001, or 1CT 3 , therefore -0007 = 10- 4+adeclmaK 

From observing the above results we infer: 

(1) That the characteristic of the logarithm of a number less 
than unity is negative. 

(2) That the characteristic of the logarithm of a number less 
than unity is one more than the number of zeros between the 
decimal point and the first significant figure. 

149. How to write the Logarithm of a Number less than 
Unity. 

Find log -067. 

By the above the characteristic is — 2, and from the tables 
the mantissa is -82607. 
.'. log -067= -2+ -82607. 

Since the mantissa is always positive we could not correctly 
write this as — 2-82607, for that would imply that the whole 
quantity 2-82607 is negative. To avoid this difficulty the 
minus sign is placed immediately above the characteristic. 

.'. log -067 = 2-82607, the characteristic being read " bar" 2. 
Example 1: Divide -0432 by 82-624. 
log -0432 = 2-63548 
log 82-624 = 1-917111 
Diff. of logs = 2- 63548 
1-917111 



4- 718369 = 4- 71837 to 5 places. 

Here note that we are subtracting the greater quantity from 
the less, therefore in obtaining the 4 we use the law for algebraic 
subtraction, i.e., change the sign of the lower line and add. 

Antilog of 4-71837= -000522844= -00052284 approx. 



236 MATHEMATICS FOR TECHNICAL SCHOOLS 

Example 2: 

to . rt , f 36-215X-0724 

FlndtheValue0f -0027X936-- 

log 36 • 215 = 1 • 55889 log • 0027 =3-43 136 

log -0724 =2-85974 log 936 =2-97128 



•41863 (a) -40264 (6) 

(a) -(6) = -41863 
•40264 



•01599- 
Antilog of • 01599 = 1 • 0374 1 = 1 • 0374 approx. 

Exercises LXXXVII. 

Employ logarithms to find the value of : 

1. 43-752H-8-75. 472-86X15-8X10" 3 

2. • 0752 -i- -648. • 0728 X • 63 X 10 2 

26 • 584 X- 075 728 • 43 X -00625X19 . 

" 8-359 -0946X1-0009 

4. 408-039-^3423-08. 

150. Logarithm of a Power. 
From tables log 2 = • 30103. 
.'. 2 = 10- 30103 - 

(O) 2 = (\ n- 30103 ) 2 = 1Q -60206. 

In the above we observe that the log of 2 2 is twice the log 
of 2, therefore to find the value of 2 2 , we would find the log 
of 2, double it and find the antilog of the result. 

Thus, log 2= -30103, 

twice log 2= -60206. 

Antilog -60206 = 3-99996 = 4 (nearly). 

Again, log 3 =-47712. 
.-. 3 = 10 47712 - 

. Q4 = /-1Q.47712\4_ JQ1.90848. 



LOGARITHMS 237 

Here we observe that the log of 3 4 is four times the log of 
3, therefore to find the value, of 3 4 , we would find log 3, take 
four times it, and find the antilog of the result. 

Thus, log 3 =-47712, 

four times log 3 = 1 • 90848. 

Antilog 1 • 90848 = 80 • 9988 = 8 1 (nearly) . 

Further log 9 = • 95424. 

.-. 9 =10 95424 - 

.'. 9i = (10 95424 )i = 10- 47712 - 

Here we observe that log 9* is one-half the log 9, therefore 
to find the value of 9*, we would find log 9, take one-half 
of it, and find the antilog of the result. 

Thus, log 9= -95424. 

$ log 9= -47712. 

Antilog -47712 = 3-00004 = 3 (nearly). 

From these examples we infer: — To obtain any power of a 
number multiply its logarithm by the exponent of the power and 
find the antilog arithm of the result. 

Example 1 : 

Find value of (-026) 3 . 
Letx=(-026) 3 . 

Then by above log x = 3 log • 026. 
= 3 (2-41497). 

Here we have to multiply a logarithm by 3, the mantissa 
being positive and the characteristic negative. We should 
first multiply them separately, giving 6+1-24491, and after- 
wards combine giving 5-24491. 

.'. log x = 5- 24491. 

x = • 0000175754 = - 000017575 approx. 



238 MATHEMATICS FOR TECHNICAL SCHOOLS 

Example 2: 

Find value of (-026)*. 

Let log s = (-026)*. 

then log x = \ log -026. 
= \ (2-41497). 

The same difficulty is presented here as in the preceding 
example, only we have to divide by 3 instead of multiplying. 
We, therefore, write $(2-41497) as £(3 + 1-41497), the object 
being to make the negative part so that the 3 will divide it 
evenly. 

1(3 + 1-41497) =1-471656. 

.*. log x = 1-471656 = 1-47166 to 5 places. 
x= -296251= • 29625 approx. 

Example 3: 

Find the value of 

Letx= (Fo5y 

then log x = log 1 — 6 log 1-05. 
= 0-6(-02119). 
= -•12714. 
Since the mantissa must always be positive we must now 
change — • 12714 to a number having a positive mantissa. 

Thus,- -12714=1 + 1- -12714. 

= 1/87286. 
.-.log* =1-87286. 

x =-746214= -74621 approx. 

151. Solution of an Exponential Equation. 
If 3 Z =148, find a:. 

This is an exponential equation, the unknown quantity being 
the exponent. 






LOGARITHMS 239 



x = 



Here x log 3 = log 148 

l og 148 
log 3 

^ 2-17026 
X -47712 
.'. log x = log 2 -17026 -log -47712. 
= -336512- 1-678628. 
- -657884= -65788 to 5 places. 
x = 4 • 54853 = 4 • 5485 approx. 

Exercises LXXXVTII. 

Find the cube root of the following numbers: 
1- 27-27. 3. -00069. 5. 437-72. 

2. -08765. 4. -7248. 6. 9281-4. 

Find the numerical value of: 
7 /^i 2 !! 1 Q 41X0015)* 

■ t25 - 34/ ' 9 - ~im~' 

10 (46-43) 10 X(-0348) 12 
•0275 1* 1-245X163 

V T 0l83j ' 11. 15 3 - 2 - 

12 (-19)*X(-19)*X(-19)*X264 2 y _1_ 
(•0418) 2 X(-4365) 3 Xv/472 10 6 ' 
3 (34-96) 14 X(165-3)- 34 1 
' (-258)- 7 X(-045) 65 10 12 * 



14. V& + &. 
1 



15 



(1-05) 8 (1-05) 



20 



16. ^Ux J 



(1-06) 8 ~ 1-05) 15 
Find the value of x in the following: 
17. 13* = 432. 18. 6 x = 25-2. 19. 15* =5. 



240 MATHEMATICS FOR TECHNICAL SCHOOLS 

Employ the formula for the area of a triangle in terms of 
its sides to find the area of the following traingles : 

20. 36-4 yd., 21-3 yd., 26-5 yd. 

21. 16-48", 23-39", 31-18". 

22. 2500 links, 3500 links, 4000 links (area in acres). 

23. 27-6 chains, 19-5 chains, 14-3 chains (area in acres). 

24. Find the length of the perpendicular drawn from A on 
BC in the triangle ABC, if a = 700', 6 = 670', c = 527-2'. 

25. The sides of a triangle are 43-6", 51-8", and 62-4". 
Find the side of an equilateral triangle of equal area. 

26. Find the area of a circle whose radius is 72-46". 

27. A circle has a radius of 43-46". Find the radius of 
the concentric circle which divides the first circle into two 
equal areas. 

28. Find the diameter of a circle whose area is equal to 
that of an equilateral triangle on a side of 18". 

29. Find the number of gallons in a cubical cistern, each 
side of which measures 18-6' (1 gal. =277-274 cu. in.). 

30. The water contained in a cubical cistern, each edge of 
which measures 5', is found to lose by evaporation -03 of its 
volume in a day. If the total loss be due entirely to evapora- 
tion, find how many gallons will be left in the cistern at the 
end of 9 days, assuming it to be full at the outset. 






CP 



CHAPTER XVI. 
MENSURATION OF SOLIDS. 

152. We have already found the surfaces and volumes of 

various rectangular solids. We will now proceed to deal with 
some of the more specialized forms of solids. 

If the block in Figure 103 has the dimensions indicated, we 
can find the area of the sides, i.e., the lateral surface by finding 
the area of each lateral face and adding the 
results. Thus the area of the front and back 
faces = 6"X18"X2 = 216 sq. in., the area of 
the two side faces = 4"X18"X2 = 144 sq. in., 
giving a total lateral area of '360 sq. in. 

The same result might have been obtained 
by first finding the perimeter of the base and 
multiplying this result by the height. Thus 
perimeter of base = 6" + 6" + 4" + 4" = 20". 
.'. lateral surface = 20" X 18" =360 sq. in. 
Further in finding the volume of this solid 
we multiplied together the three dimensions — 
length, breadth and thickness. Thus volume 
= 18"X6"X4" = 432 cu. in. The same result 
might have been obtained by first finding the area of the base 
and then multiplying this area by the height. Thus area of end 
= 6"X4" = 24 sq. in. and volume = 24X18" = 432 cu. in. 

153. The Prism. A prism is a solid whose sides are paral- 
lelograms and whose top and bottom are parallel to each other. 

In Figure 104 we have represented a number of prisms 
each complying with the conditions in the definition. 

A prism is called triangular, rectangular, pentagonal, etc., 
according as the base is one or other of these polygons. 

241 



- — €>"—»■ 

Fig. 103 



242 



MATHEMATICS FOR TECHNICAL SCHOOLS 



To find the lateral surface of any of the prisms below we 
would proceed as in Figure 103, i.e., multiply the perimeter of 
the base by the height. Thus if p be the perimeter of the base 
and h the height, the area of the lateral surface of the 
prism = ph. 



a 





Fig. Kit 

To find the volume of any one of the above prisms we would 
as in Figure 103 multiply the area of the base by the height. Thus 
if b be the area of the base and h the height, the volume of 
the prism = bh. 

If we wish to find the area of the total surface of a prism, 
we would add the areas of the two ends to the area of the lateral 
surface. 

Exercises LXXXIX. 

1. Measure the various prisms in the laboratory. Make draw- 
ings in your laboratory book and find total area and volume. 

2. The internal dimensions of a box, without a lid, are 
length 8', breadth 3', depth 2'. Find the cost of lining it with 
zinc at 40c. a sq. ft. 

3. A rectangular tank, 13' 6" in length by 9' 9" in breadth, 
is full of water. How many gallons of water must be drawn 
off to lower the surface 1 "? 

4. How many sq. ft. of metal are there in a rectangular 
tank, open at the top, 12' in length 10' in breadth and 8' deep? 

5. A prism whose base is a regular pentagon with a side 
of 9|" is 25^" in height. Find its total area and volume. 

6. A rectangular tank is 11|" long, 14 \* wide, and 10" 
deep. Find the number of gallons it contains when filled 
with water within an inch of the top. • 



MENSURATION OF SOLIDS 



243 




154. The Cylinder. A cylinder is a solid whose lateral 
surface is curved and whose bases are parallel to each other. 

To find the lateral surface of a 
cylinder. If we roll a cylinder on 
a sheet of paper until it has made 
one complete revolution, we observe 
that the area of the paper touched 
by the cylinder is a rectangle 
whose length is equal to the cir- 
cumference of the cylinder, and 
whose breadth is equal to the height 
of the cylinder. From this experi- 
ment we infer that the lateral surface 
of a cylinder = the circumference of 
base multiplied by the height. Fig. 105 

Therefore with the notation in the figure the area of the 
lateral surface =2irrh 

= irdh (d = diameter) . 

To find the volume of the cylinder. 
If the cylinder in Figure 106 be cut 
into a number of triangular prisms 
as indicated, we can find its volume 
by adding together the volumes of the 
prisms. Since the volume of a tri- 
angular prism is found by multiplying 
the area of the base • by the height, 
the volume of all the prisms, i.e., of 
the cylinder, may be found by multi- 
plying the sum of the areas of the bases 
by the height. Therefore the volume 
of a cylinder = area of base multiplied 
by the height, or V = -rrr 2 h= -785±d 2 h. 

Example : 

A cylindrical tank open at top is 6' 
high and has a diameter of 3'. Find (1) the cost of lining 





Fig. 106 



244 



MATHEMATICS FOR TECHNICAL SCHOOLS 



with galvanized iron at 20c. a sq. ft., (2) its capacity in 
gallons. 

Area of lateral surface =rX3X6 = 18T sq. ft. 



Area of bottom 


= tt(|) 2 sq. ft. «= 2 • 25 * sq. ft 


.'. total area 


= tt(18+2-25) sq.ft. 




= 7r(20-25) sq. ft. 


.•. cost 


= 7r(20-25)X20 = $12-72 


Volume 


= *"(f) 2 X6 cu. ft. 


Capacity 


= 7r(f) 2 x6X6-232gal. 




= 264-31 gal 




Exercises XC. 



1. Measure the various cylindrical models in the labora- 
tory. Make drawings in your laboratory book and obtain 
the lateral area and volume in each case. In the case of the 
iron and steel models find their weights from knowing their 
volumes. Check by weighing. 

2. Fill in the omitted entries in the following cylinders: 



No. 


Diameter 


Height 


Circ. at Base 


Area of Base 


Lateral Area 


Volume 


1 


5" 


H" 










2 


8" 


7" 










3 


1" 








28 sq. ft. 




4 




3' 




154 sq.ft. 






5 




V 








616cu.ft. 


6 




8' 


44' 









3. Find the weight of a steel shaft 2" in diameter and 12' 
long. 

4. A tank car is 33£' long and 8£' in diameter. How many 
gallons of oil will it contain? 

5. Find the cost of painting the inside of an open cylindrical 
tank 10' in diameter and 15' high at 20c. a sq. yd. 



MENSURATION OF SOLIDS 



245 



6. A cylindrical vessel partly filled with water is 8" in 
diameter. A steel crane hook is immersed in the vessel and 
the surface of the water is raised 2". Find the weight of the 
crane hook. 





Fig. 107 

155. The Hollow Cylinder. The total surface of the hollow 
cylinder in Figure 107 would consist of the outside lateral 
surface, the inside lateral surface, and the two rims. 
The outside lateral surface = ttX8X 18 = 144 * sq. in. 
The inside lateral surface = tt X 6 X 18 = 108tt sq. in. 
The area of the rims = ttX7X1X2 = 14tt sq. in. 

The total lateral surface = 266 ■* sq. in. =835-68 sq. in. 
The volume of the hollow cylinder would he the area of the 
base multiplied by the height. 

Area of the base, i.e., the area of the ring in Figure 107 
= tX7X1 = 7tt sq. in. 
.'. the volume = 7ttX 18 = 395 -84 cu. in. 

Exercises XCI. 

1. Measure the hollow cylindrical models in the laboratory. 
Make drawings in your laboratory book and calculate the total 
surfaces and volumes. 

2. Find the whole surface of a hollow cylindrical pipe, 
open at the ends, if the length is 8", the external diameter 
10" and the thickness 2". 



246 



MATHEMATICS FOR TECHNICAL SCHOOLS 



3. An iron roller is in the shape of a hollow cylinder whose 
length is 4', external diameter 2' 8" and thickness \" . Find 
its weight if a cu. ft. .of iron weighs 486 lb. 

4. A portion of a cylindrical steel shaft casing is \2\' in 
length, \\" thick, and its external diameter is 14". Find its 
weight. 

5. Find the weight of a lead pipe 8' long, external diameter 
8", internal diameter 7", assuming that the weight of the two 
flanges is equivalent to one foot length of pipe. 

6. Find the weight of a hexagonal cast-iron nut 1" to the 
side, \" thick, inside diameter f ". 

156. The Right Cone. A cone is a solid whose base is a 
circle and whose sides taper uniformly to a point directly 
over the base. 




Fig. 108 



Fig. 109 



Lateral Surface of a Cone. If a piece of paper be wrapped, 
without crumpling or tearing, around the lateral surface of a 
cone (Figure 108) and cut along the edge of the base and the 
line AB, and then folded out, the paper will be a sector of a 
circle (Figure 109). 



MENSURATION OF SOLIDS 



247 



The radius AB of this sector is equal to the slant height of 
the cone and the length of the arc BD is equal to the 
circumference of the base of the cone. The area of a sector 
of a circle has previously been found to be equal to fare 
X radius. Therefore the lateral surface of a cone = \cir- 
cumference of the base multiplied by the slant height, or with 
the notation of the figure, the lateral surface of a cone 
= 2irr X hs = *d X \». 

Perform the experiment suggested above. Make drawings 
and write conclusions in your laboratory book. 





Fig. 110 



The Volume of a Cone. Take two vessels, one conical and 
the other cylindrical, having the same height and radius of 
end. (Figure 110). 

If we fill the conical vessel with water and empty it into the 
cylindrical one, we find that it will take three fillings of the 
conical vessel to fill the cylindrical one. 

We, therefore, infer that when the vessels are related as 
in the above illustration, the volume of the cone is one-third 
that of the cylinder. 



248 MATHEMATICS FOR TECHNICAL SCHOOLS 

But the volume of the cylinder = area of base multiplied 
by the height. Therefore the volume of a cone = area of 
base X3 perpendicular height, or V = irr 2 X\h = \ir r 2 h. 

Note. — Height means perpendicular height, unless otherwise 
stated, but in the formula for the volume of a cone we should 
state "perpendicular" height to distinguish from "slant" 
height in the formula for the lateral surface. 

Perform the experiment suggested above. Make drawings 
and write conclusions in your laboratory book. 

Example : 

A conical tent has a diameter at the base of 14' and a height 
of 7'. 

Find (1) the number of sq. yd. of canvas in the tent. 

(2) the number of cu. ft. of air space. 
Slant height of cone = V7 2 +7 2 = 9-89' 
Number of sq. yd. =7rX14X 9 -^p X£ 

= 24-17 
Airspace = 7rX7'X7Xf = 359- 19 cu. ft. 

Exercises XCII. 

1. Measure the various conical models in the laboratory. 
Make drawings in your laboratory book and calculate lateral 
surfaces and volumes. Find the weights of iron models from 
knowing their volumes. Check by weighing. 

2. A piece of paper in the form of a circular sector, of 
which the radius is 8" and the length of the arc 12", is formed 
into a conical cap. Find the area of the conical surface and 
the base of the cone. 

3. Find the weight of a cast-iron cone, diameter of base 
7" and height 15". 

4. Find the weight of petroleum in a conical vessel, diameter 
of the base 14", height 10", specific gravity of petroleum -87. 

5. The interior of a building is in the form of a cylinder of 
20' radius and 15' in height. A cone surmounts it, radius of 
base 20' and height 8'. Find (a) the cost of painting the 
interior at 20c. a sq. yd., making no allowance for openings, 
(6) cubic feet of air space in the building. 



MENSURATION OF SOLIDS 



249 



6. How many yards of canvas 27" wide will be required to 
make a conical tent 7 yd. in diameter and 10' high? 

157. The Pyramid. A pyramid is a solid whose sides are 
triangles and whose base is any figure bounded by straight 
lines. 

In Figure 111 we have the simplest 
type of a right pyramid, the base 
being a square. 

Lateral Surface of a Pyramid. In 
Figure 111 the lateral surface consists 
of four equal isosceles triangles. 
Area of ACD = CDXhAE. 

.*. area of four faces = 4 times 
CDXhAE. 

But 4 times CD = perimeter of base, and AE = slant height 
of pyramid. 

.'. lateral surface of pyramid = perimeter of base X | slant height. 

= \ V s (V = perimeter, 5 = slant ht.). 
It may readily be shown that this formula holds where the 
base is any regular polygon. 








Fig. 112 



Volume of a Pyramid. Take two vessels one a square 
pyramid and the other a rectangular prism of the same height 



250 MATHEMATICS FOR TECHNICAL SCHOOLS 

and area of end as in Figure 112. If we fill the pyramidal 
vessel with sand and empty it into the prism, we find that 
it takes three fillings of the pyramid to fill the prism. We 
therefore infer that, when the vessels are related as above, 
the volume of the pyramid is one-third that of the prism. 

The volume of prism = Area of base multiplied by the 
height. 

.*. volume of pyramid = area of baseX^perp.height. 
or V = \Ah (A = area of base, h = height). 
Example: 

A granite pyramid 12' high stands on a square base 10' 
to the side. Find (1) cost of polishing the lateral surface at 
10c. a sq. ft. (2) weight, if 1 cu. ft. weighs 165 lb. 
Slant height = Vl2 2 +5 2 = 13' 

Lateral surface =4X10X -^ sq. ft. 
Cost of polishing = 4 X10X- 1 / X -^ =$26.00 
Volume = 1X10X10X12 

= 400 cu. ft. 
Weight =165X400 
= 66,000 lb. 

Exercises XCIII. 

1. Measure the various pyramidal models in the laboratory. 
Make drawings and calculate lateral surfaces and volumes. 

2. What is the weight of a cast-iron pyramid with a square 
base 6" to a side and a height of 10"? 

3. Find the total surface of a hexagonal pyramid with a 
base 3" to the side and a slant height of 12". Find its weight 
if made of cast-iron. 

4. Find the number of cu. ft. of air space in a hexagonal 
room, each side of which is 12', and its height 18', which is 
furnished above with a pyramidal roof 9' high. Find also 
the cost of painting the interior at 25c. a sq. yd., making no 
allowance for openings. 

5. A pyramid has a square base each side of which is 2-48", 
and the pyramid has equilateral triangles for sides. Find 
its volume. 



MENSURATION OF SOLIDS 



251 



Frustum of Cone or Pyramid. A frustum of a cone or 
pyramid is the part contained between the base and a plane 
drawn parallel to it. 

Lateral Surface of Frustum 
of Cone. aABb in Figure 113 
may be considered as a tra- 
pezium, ab and AB being the 
parallel sides and either aA 
or bB representing the per- 
pendicular distance between 
the parallel sides. 

If we consider this figure 
as being bent around until 
a coincides with b and A 
with B, it would take the 
form of a frustum of a cone, 

the parallel sides of the trapezium becoming the circumferences 
of the ends and the perpendicular distance between the 
parallels becoming the slant height of the frustum. 




Fig. 113 




Fig. 1U 



Since area of trapezium = Sum of parallel sides multiplied 
by | perp. distance between them. 



252 



MATHEMATICS FOR TECHNICAL SCHOOLS 




.'. lateral surface of frustum of cone = sum of circumferences 
of ends X§ slant ht. 

or, Lateral surface = \ (C+c)S. 
= tt (R+r)S. 
Lateral Surface of Frustum of Pyramid. If we consider the 

frustum of a pyramid in 
Figure 115, we observe that 
its lateral surface is made 
up of four equal trapeziums. 
Area of the face cCDd = 
{cd+CD)\qQ. 

.'. area of the four faces = 
±(cd + CD)\qQ. 

But 4 (cd+CZ))= Sum of 
perimeters of ends and qQ = slant height of frustum. 

.*. lateral surfaces of frustum 
of / pyramid=sum of perimeters 
of ends X \ slant ht. = \ 
(Pi + Pa) S (P x and P 3 being 
perimeters and S slant height). 

Volume of Frustum of Cone 
or Pyramid. In Figure 116 
from similar triangles Ocb and 
x _ r 
x~+h~R' 

.'. x-\-h = 



Fig. 115 



OCB we have 
hr 



- • x= » 

K— r 

Volume of whole cone 

= l T /?2 IlR 

1 ' R-r 
Volume of small cone 
, , hr 




volume of frustum = l^h 



Fig. 116 



R 3 -r- 



R-r 

i7rh{R 2 + Rr+r 2 }. 



MENSURATION OF SOLIDS 



253 



If A represents area of large end and a area of small end, 
then A — R 2 and a = Trr 2 . 

.". volume of frustum ^AA -ha+s/Aa}- 

o 

Work through a similar proof to show that the volume of a 
frustum of a pyramid is the same as the above. 

Example: 

A vessel in the form of a 
frustum of a cone has the 
following dimensions: Depth 
16", diameter of large end 
12", diameter of small end 
8". Find (a) its lateral 
surface (b) its capacity in 
gallons. 

In the rt.-angled triangle 
ABC, ^C=Vl6 2 +2 2 
= 16-12". 

Lateral surface 



= (7rl2+7r8) 



16-12 



= 20^x8-06 = 506-43 sq. in. 
Volume 

= JUL {7r6 2 + ir4 2 +\A6 2 ir4 2 } 

= -V L {^6 2 +7r4 2 +7r6X4} 
= ^7r{6 2 +4 2 +24}- 




Fig. 117 



= - 1 / 7T 76 cu. in 



167 



76 



Capacity in gallons = ^- X 2? 



4-59. 



Exercises XCIV. 

1. Measure the frustum models in the laboratory. Make 
drawings in your laboratory book and calculate lateral sur- 
faces and volumes. 

In the case of the iron and steel models find weights from 
knowing their volumes. Check by weighing. 



254 



MATHEMATICS FOR TECHNICAL SCHOOLS 



2. Find the lateral surface of the frustum of a pyramid, 
perpendicular height 6", and a square base, side 6", the side 
of the upper square being 1". 

3. A tapered piece of cast-iron 2' long is 8" in diameter at 
one end and 12" in diameter at the other; find its weight. 

4. A piece of steel 16" long is 4" in diameter at the large 
end. The taper is a Brown and Sharpe — \" to 1'; find its 
weight. 

5. Find the volume of a steel pin 8" long, diameter of small 
end 2", the taper being a No. Morse — f " to 1'. 

6. Two buckets, one cylindrical of 7" diameter, the other 
a frustum of a cone with the diameters of its ends 6" and 8" 
are of the same depth, 9". Find the difference in their volume. 



158. The Sphere. A sphere is the geometrical name for a 
round or ball-shaped solid. 





Fig. 119 



Area of Surface of Sphere. It has been found by measure- 
ment that the surface of a sphere is equal to the lateral 
surface of a cylinder of the same diameter and height, as 
illustrated in Figure 119. 

The circumference of the cylinder is 2nr and its height 2r, 
hence area of surface of sphere = 27rrX2r = 4tjt 2 . 



MENSURATION OF SOLIDS 



255 



An approximate idea of this relation may be obtained by 
the following experiment : 

Insert a nail in the centre of the curved surface of a hemi- 
sphere. Fasten the end of a cord to this nail and wrap it with 
the object of completely covering the curved surface. 

Next insert a nail in the centre of the flat face and wrap the 
cord with the object of completely covering the flat surface. 
It may then be observed that the length of cord required to 
cover the flat surface is only half that required to cover the 
curved surface. 

But area of flat surface =7rr 2 . 

.*. area of curved surface =27rr 2 . 

.\ total surface of sphere = 47rr 2 . 

Since 4tit 2 = 7r(2r) 2 .\ area of surface of sphere in terms of 
diameter = 7rD 2 . 

Volume of Sphere. The sphere may be considered as made 
up of a number of pyramids, as in- 
dicated in Figure 120, whose bases 
together form the surface of the sphere 
and whose apexes meet at the centre. 

Since the volume of a pyramid equals 
area of base multiplied by ^ perp. ht. 
( = radius of sphere), .\ volume of sphere 
= Surface X i radius. 
= 4;rr 2 X| r. 




Fig. 120 



-* 



Volume in terms of diameter = -5236Z) 3 . 
Example : 

Find the surface and weight of a cast-iron ball, radius 5". 

Surface = 4tt5 2 = 314- 16 sq. in. 

Volume =|7r5 3 = f Ti-125 cu. in. 

Weight =|7rl25X-26 = 136-13 1b. 



256 MATHEMATICS FOR TECHNICAL SCHOOLS 

Exercises XCV. 

1. Measure the spherical models in the laboratory. Cal- 
culate areas and volumes. 

2. Secure cylinder and sphere related as in Figure 119. 
After placing sphere in cylinder, fill the remaining space with 
sand. Remove sphere and replace the sand. By estimating 
the part of the cylinder now occupied by the sand derive the 
formula for the volume of the sphere. 

3. Find the number of yards of material, 27" wide, necessary 
to make a spherical balloon 12' in diameter. 

4. Find the weight of a ball composed of a cast-iron sphere 
4" in diameter, covered with a layer of lead 1" thick. 

5. Find the weight of a hollow cast-iron sphere, internal 
diameter 2\", thickness \" . 

6. How many ounces of nickel would be used in plating 
a ball 3" in diameter, to a depth of ^-"? (1 cu. in. nickel 
weighs 5-14 oz.). 

Segment of a Sphere. A segment of a sphere is the part 

cut off from a sphere by a plane. 

Lateral Surface of a Segment. If we roll a sphere on a sheet of 

paper, and keep in mind that the area 

of the surface is equal to that of a 

cylinder with radius of base equal to the 

radius of the sphere and height equal to 

the diameter of the sphere, we could infer 

fig. 121 that the surface traced out by any 

segment is equal in area to a rectangle 

having the circumference of the sphere for length and the 

height of the segment for width. 

.'. lateral surface of segment 

= 2ttRx1i. 

= 2ttM. 

The volume of the segment in Figure 121 (less than a hemi- 

vJir^ ttJi^ 
sphere) is given by the formula : V = —= — H -jr~ • 

Z o 

Zone of Sphere. A zone of a sphere is the part cut off from 
the sphere between two parallel planes. 




MENSURATION OF SOLIDS 



257 



Lateral Surface of Zone 
the area traced out when 
paper would be equal in area to a 
rectangle having the circumference of the 
sphere for length, and the thickness of 
the zone for breadth. 

.". lateral surface of zone = 2irRh. 

The volume of the zone in Figure 122 
is given by the formula: 



As in the segment of a sphere 
rolled on the 





S03-JZ^ 


1* 


f T T \ 


1= 


z=zfi53T— 



Fig. 122 



Sector of Sphere. 




Fig. 123 



A sector of a sphere consists of a segment 
and a cone whose bases are coincident, 
the apex of the cone being at the 
centre of the sphere. 

The surface of the sector would be 
equal to the surface of the segment 
plus the surface of the cone. 

The volume of the sector 



is given 



by the formula : 



V = l{r>(h+2R) +/* 3 }- 



159. Bead. Volume 
sphere is pierced by a 



remaining 
cylindrical 



when 
solid. 



Volume of bead as 
bead). 



shown = - n -(h = ht. of 
6 




Fig. 124 



Exercises XCVI. 

1. The silk covering of an umbrella forms a portion of a 
sphere of 3^' in diameter, the area of the silk being 14 1 sq. 
ft. Find the area sheltered from vertical rain when the handle 
is held upright. 

2. A sphere of diameter 24' is placed so that its centre is 
37' distant from the observer's eye. Find the area of that 
part of the sphere's surface that is visible to the observer. 



258 



MATHEMATICS FOR TECHNICAL SCHOOLS 



3. A cylindrical tank is 8' long and 2\' in diameter. The 
ends are spherical segments whose centre of curvature projects 
6" beyond the base of the segment. Find the total surface 
and volume of the tank. 

4. If the diameters of two circles of a spherical zone are 
12 " and 4", and the thickness of the zone 6", find its total 
surface and volume. 

5. In the sector of a sphere of radius 10", the height of the 
segment is 4"; find the volume of the sector. 



160. Solid Ring. 




Ftg. 125 



Examples of solid rings are found in 
anchor rings, curtain rings, etc. It will 
be observed that any cross-section of 
such a ring will be a circle, so it may be 
considered as a cylinder bent around in a 
circular arc until the ends meet. The 
mean length of the cylinder will be 2irR. 

.'. with notation of figure: 

Surface = 2 rrr x 2->rR = 4:7r 2 rR. 
Volume = Trr 2 X 2*R = 2* 2 r 2 R. 



161. Wedge. A wedge, as shown, is a solid contained by 
five plane faces; the base is a rectangle, the two ends are 
triangles, and the two remaining 
faces are trapeziums having a 
common side, called the edge, 
which is parallel to the base. 

The surface of the wedge is 
found by calculating separately 
the area of each of the faces. To 
do this, the slant heights of the 
faces, or means of finding them, must be given. 

The volume of the wedge is given by the formula: 




Fig. 126 




MENSURATION OF SOLIDS 259 

162. Prismoid — An irregular-shaped solid having five or 
more flat or plane faces, two of which are 
parallel. 

Volume of prismoid given by the /,£ 

formute: V = ^{A + B-\-U1\, 
o 

where A and B are the areas of the FlG 127 

parallel faces, M the area of a section 

half-way between them, and h the height. 

Exercises XCVII. 

1. Find the weight of a brass wedge, whose height is 5" 
and edge 4", the base being a rectangle which measures 8" 
by 6" (1 cu. in. = -3 lb.). 

2. How many tons of earth are removed in excavating a 
trench of which the top and bottom are rectangles? At the 
top it is 400' long by 18' wide, and at the bottom it is 350' 
long by 15' wide. The bottom is horizontal and the depth 
12', (given 1000 cu. ft. earth weighs 40 tons). 

3. In a cast-iron wheel the inner diameter of the rim is 
2' and the cross-section of the rim is a circle of 6" radius; find 
the weight of the rim. 

4. The cross-section of the rim of a cast-iron fly-wheel is 
a rectangle 8" by 10". If the mean diameter is 10', find the 
weight of the rim. 

5. A wedge-shaped trench is 40 yards long at the top and 
8' wide; the length of the bottom edge is 32 yards and the 
depth is 10'. How many cu. yd. of earth have been excavated? 

6. The cross-section of the rim of a fly-wheel is a rectangle 
6" by 8", the shorter dimension being in the diameter of the 
wheel. The wheel is 22' in outer diameter; find the weight 
if the specific gravity of the material is 7 • 2. 

Miscellaneous Exercises XCVIII. 
(1 cu. in. cast-iron = -26 lb. 1 cu. in. steel = -2834 lb.) 

1. A square bar of wrought-iron 12' long, weighs 80 lb. 
What is the size of the end? 

2. A closed cast-iron tank is 3' long, 2f wide, and 2§' 
deep, outside measurements. If the material is \" thick, 
find the weight of the tank. 



260 MATHEMATICS FOR TECHNICAL SCHOOLS 

3. The rain which falls on a roof 22' by 36' is conducted to 
a cylindrical cistern 8' in diameter. How great a rainfall 
would it take to fill the cistern to a depth of 7|'? 

4. Water is poured into a cylindrical reservoir 20' in 
diameter, at the rate of 300 gallons per minute. Find the 
rate, in feet per minute, at which the water rises in the 
reservoir. 

5. The internal diameter of a cylinder, open at the top, is 
1|', and its weight is 180 lb.; when filled with water it weighs 
2000 lb. ; find the depth of the cylinder. 

6. Find the weight of a copper tube f " outside diameter, 
•05" thick, and 5' 10" long. 

7. A steel bar whose cross-section is a regular hexagon 1" 
to the side, is 8' in length. Find its weight. 

8. A trough whose cross-section is an equilateral triangle 
8" to the side contains 30 gallons of water; how long is it? 

9. A boiler has 275 tubes, each 19' 3" long and 2f" in 
diameter. What is the total heating surface of the tubes? 

10. Find the capacity in gallons of a conical vessel 15" in 
diameter and 2' in slant height. 

11. A conical tent covers an area of 154 sq. ft. and is 6' 
in height. How many sq. yd. of canvas does it contain? 

12. What is the volume of a cylindrical ring having an 
outside diameter of 6|", an inside diameter of 5 A", and a 
height of 5f "? 

13. Water flows at the rate of 20' per min. from a cylindri- 
cal pipe -25" in diameter. How long would it take to fill a 
conical vessel, whose diameter at the surface is 10" and depth 
9"? 

14. The external diameter of a hollow steel shaft is 20", 
and the internal diameter 12". Find the weight of 20' of 
this shafting. 

15. From a cylinder whose height is 8", and diameter 12", 
a conical cavity of the same height and base is hollowed out. 
Find the whole surface of the remaining solid. 

16. Find the cost of polishing the lateral surface of a 
pyramid 6' 5" high, standing on a square base 6' to the side, 
at the rate of 20 c. a sq. ft. 

17. How many gallons of water will be discharged per 
min. from a 4" pipe if it flows at the rate of 300' per minute? 



MENSURATION OF SOLIDS 261 

18. The cross-section of a water pipe is a regular hexagon 
whose side is 1". At what rate, in feet per min., must the 
water flow through the pipe in order to fill in one hour a 
cylindrical tank the radius of whose base is 16'' and whose 
depth is 5'? 

19. The base of a prism whose altitude is 15" is a quadri- 
lateral whose sides are 10", 18", 12", 16", the last two forming 
a rt. angle. Find its volume. 

20. A tower whose ground plan is a square on a side of 
30', is furnished with a pyramidal roof 8' high. Find the 
cost of covering the roof with sheet-iron at 25c. a sq. ft. 

21. A steel bar whose cross-section is an equilateral triangle 
1\" to the side is 8' long; find its weight. 

22. A cylindrical granite pillar 10' high and 30" in diameter, 
is surmounted by a cone 2\' high. Find the weight of the 
whole if a cu. ft. of granite weighs 165 lb. 

23. How many cu. in. are there in a hexagonal blank 
nut -5" to a side and f" thick? 

24. It is desired to make a conical oil can with a base 5" 
in diameter to contain \ pint; what must be the height? 

25. A piece of cast-iron has a B. & S. taper — -\" to 1'. It 
is 10" long and the diameter at the large end is 3-5"; find its 
weight. 

26. Find the height of a pyramid, of which the volume is 
625 cu. in. and the base a regular hexagon 12" to the side. 

27. The perpendicular height of a square chimney is 
150' 3". The side of the base measures 12' 6" and the side 
at the top 6' 3", the cavity is a square prism whose side meas- 
ures 3' 9". How many cu. ft. of masonry in the chimney? 

28. A circular disc of lead, 3" in thickness and 12" diameter, 
is converted into shot, each -05" in radius. How many shot 
does it make? 

29. The interior of a building, in the form of a cylinder 
of 15' 0" radius and 10' 0" high, is surmounted by a cone whose 
vertical angle is a rt. angle. Find the area of the surface 
and the cubical contents of the building. 

30. A square building 20' 0" to the side has a hip roof in 
the form of a pyramid. The peak of the roof is 10' above 
the plate level and the rafter heel is 2'; find the cost of roofing 
with shingles, laid 4|" to the weather, material and labour 
costing $12 a square of shingles. 



262 MATHEMATICS FOR TECHNICAL SCHOOLS 

31. The base of a cone is an ellipse, major axis 4", minor 
axis 2", height 6". Find the volume. 

32. A quart measure is 8" in height. Find the diameter 
of its base. 

33. Find the weight of a log 40' long, 4' 6" in diameter at 
one end and 30" in diameter at the other, the specific gravity 
of the wood being -78. 

34. A piece of copper 6" long, 2" wide, and \" thick, is 
drawn out into a wire of uniform thickness and 100' long. 
Find the diameter of the wire in mils. 

35. A conical vessel 7|" deep and 20" across the top is 
completely filled with water. If sufficient water is now 
drawn off to lower the surface 6", find the area of the surface 
of the vessel thus exposed. 

36. A cylinder 2" in diameter and 8" in height contains 
equal volumes of mercury, oil and water. If the specific 
gravity of the mercury be 13-6, of oil »92, find the total 
weight of contents. 

37. The radii of the internal and external surfaces of a 
hollow spherical shell of metal are 10" and 12" respectively. 
If it is melted down and the material formed into a cube, 
find the edge of the cube. 

38. An automobile gasoline tank has an elliptical cross- 
section 9" by 15" and is 3' long. How many gallons of gasoline 
will it hold? 

39. A hemispherical basin holds 2 gallons. Find its 
internal diameter. 

40. If 30 cu. in. of gunpowder weigh 1 lb., find the internal 
diameter of a spherical shell that holds 15-4 lb. 



CHAPTER XVII. 
RESOLUTION INTO FACTORS. 

163. When a quantity is the product of two or more quantities, 
each of these is called a factor of the quantity, and the finding 
of these quantities is called factoring the quantity. 

Thus, 6 is the product of 3 and 2, therefore 3 and 2 are called 
the factors of 6. 

Further 2xy is the product of 2, x and y, therefore 2, x and y 
are called the factors of 2xy. 

164. First Type. If we wish to find the value of 6X4+6 X3, 
we could find the value of 6X4 = 24, then 6X3 = 18, and then 
add the results giving 42. 

We might also write as follows:— 6X4+6X3 = 6(4+3) = 
6X7=42. 

The following examples will illustrate this algebraically: 

1. bx-\-by-\-bz = b(x-{-y-\-z). 3. 4a; 2 — 16x?/ = 4.r(x — 4^). 

2. 2^+4 = 2(?/+2). 4. a6+ac+a = a(6+c + l). 
Frequently an expression may be resolved into factors under 

this type by arranging in groups which have a compound 
factor common. 

Thus, factor x 2 — ax+bx — ab. 

By taking x out of the first and second and b out of the third 
and fourth we may write as follows: — x {x — a) +6 (x — a). 

We have now the sum of two products with x — a in each, 
therefore we may write as {x — a)(x+6). 

Examples: 

1. Factor a 2 +ab+ac+bc. 2. Factor 12a 2 -4a6-3ax 2 +6x 2 . 

= a(a+6)+c(a+6). =3a (4a-x 2 ) -&(4a-z 2 ). 

= (a+b)(a+c). =(4a-z 2 )(3a-&). 

263 



264 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Factor: 

1. ax — a 2 . 

2. x 2 -3ax. 

3. 5x 3 -15x 2 y. 

4. 8a 3 -16a&. 

5. 21-56x. 

6. -ay + by+cy. 

7. ax — bx — ex. 

8. 3a 2 6 2 -9a6 + 12. 

9. 14x 3 -7x 2 y4-56x?/ 2 
10. 5a 2 + 15ax+20a&. 



Exercises XCIX. 



11. ax — bx — ay-\-by. 

12. x 2 — xy+xz — yz. 

13. 3x — 3y-{-ax — ay. 

14. x 3 -x?/-2x 2 4-2?/. 

15. ab (x 2 + l)-x (a 2 +6 2 ). 

16. a 5 + a 4 + a + l. 

17. a 2 -bc-b + a 2 c. 

18. 2a 3 +6a 2 -ca-3c. 

19. x 2 + mx (m + l) + m 3 . 

20. ax+6x+a«/+6?/ — az — bz. 



165. Second Type. In the treatment of multiplication we 
found the product of two binomials asx+2andx-f-5 as follows: 

x+2 

x+5 

x 2 +2x 

+5x + 10 
a; 2 + 7x + 10* 

While the result could always be obtained by this method, 
it is important that the student should be able to write down 
the product of two binomials by inspection. In the result 
above we observe that the first term is the product of the first 
terms of the two expressions; the third term is the product 
of the second terms of the two expressions; the middle term 
has for its coefficient the sum of the numerical quantities 
(with proper sign) in the second terms of the two expressions. 

Write down the values of the following products: 



1. 


0r+4)(:r4-5). 


7. 


(x — 3a)(x + 2a). 


2. 


0-6)(x-f-2). 


8. 


(x+7ij)(x-Sy). 


3. 


(p+3)(p-6). 


9. 


(2x-5)(2x + 6). 


4. 


(r+4)(r-6). 


10. 


(3.r-l) (3x4-1). 


5. 


(x+6)(x+3). 


11. 


(2x+7y)(2x-5y) 


G. 


(p-9)(p + l). 


12. 


(2x4-a)(2x+6). 



RESOLUTION INTO FACTORS 265 

The converse problem gives us our second type and consists 
in finding the two factors if we know the product. 

Thus, factor z 2 +7z + 12. 

The second terms of the factors must be such that their 
product is + 12 and their sum +7. Hence they must both be 
positive, and it is readily seen that they must be +4 and +3. 

.-. z 2 +7x + 12 = (x+4)(x+3). 

Factor x 2 —10ax + 9a 2 . • 

The second terms of the factors must be such that their 
product is 9a 2 and their sum — 10a. Hence they must be 
— 9a and — a. 

.-. z 2 -10a 2 +9a 2 = (x-9a)(z-a). 

If we multiply 3x+4 by 2x + l we get 3x(2x + l)+4(2a; + l) 
= 6x 2 +3x+8z+4 = 6z 2 +ll:r+4. 

The converse problem is now to be considered. 

Factor 4x 2 +llz- 3. 

Here the numerical coefficients of the first terms of the 
factors must be 4 and 1, or 2 and 2, and the last terms must 
be 3 and 1. 

The possible sets (omitting the signs) are: 

4x 3, x 3, 2x 1. 

x 1, 4a; 1, 2x 3. 

Since the sign of the last term in 4x 2 +llx — 3 is minus, 
we at once decide that the signs of the last terms in the factors 
must be different, and therefore that the partial products 
must be subtracted. The second arrangement is the only 
one from which we can obtain llx, and also since the middle 
term is positive, the larger of the cross products must be 
positive. 

.*. 4z 2 +llz-3 = (x+3)(4x-l). 
Example 1: 

12x 2 -x-20=(3x-4)(4a;+5). 
Example 2: 

Factor 3a; 2 -7x+2 = (3x-l)(a;-2). 



266 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Factor and verify: 

1. z 2 +10x + 21. 

2. x 2 - IQx + 24. 

3. x 2 -4x+4. 

4. x 2 -x-2. 

5. z 2 -llx + 10. 

6. z 2 -a;-42. 

7. x 2 -3x -130. 

8. l-3x+2x 2 . 

9. z 2 +x-72. 

10. z 2 +4x-5. 

11. 5 — 4x — x 2 

12. 40-13:r+a; 2 . 

13. l-5x+6z 2 . 

14. 40-3z-x 2 . 

15. l-3x-130x 2 . 



Exercises C. 



16. 5x 2 +42z-27. 

17. 4z 2 - 16z + 15. 

18. 3x 2 -22x+7. 

19. 6x 2 -llx+3. 

20. 9z 2 -9x-28. 

21. 26x 2 -4Lr+3. 

22. 12.r 2 -17x-f-5. 

23. 5x 4 -10x 2 ?/ 2 -400?/ 4 . 

24. 2x 2 +5xy+Sy 2 . 

25. 12x 2 -2^-30.v 2 . 

26. 12x 2 -5:n/-3?/ 2 . 

27. 8x 2 +22x+9. 

28. 6x 2 -l3xy+6y 2 . 

29. 13xV-9x 4 -4?/ 4 . 

30. Ux 2 +8Sxy-6y 2 . 



166. Third Type. 



If we multiply x+y by x— y we have: 



z 2 +:n/ 



— xy — y 2 
x 2 —y 2 

Observing the above we find that, when we multiply the sum 

of x and y by the difference 
of x and y, the result is the 
difference of the squares of 
x and y. Therefore we may 
say that the difference of 
the squares of two quantities 
is equal to the sum of the 
quantities multiplied by the 
difference of the quantities. 



h« a 




-* « — 


C 














n 


t 




, 


M 

) 




i 



Fig. 128 



Geometrical Illustration: 

(a+6) (a-b) = ANHC 

= BMHC+CKFD 
= BEFD-MEKII 
= a 2 -6 2 . 



RESOLUTION INTO FACTORS 267 

Examples: 

1. p 2 -q 2 =(p+q)(p-q). 

2. 9a 2 -256 2 = (3a) 2 -(56) 2 =(3a+56)(3a-56). 

3. TR 2 -irr 2 =T(R 2 -r 2 ) = ir(R+r)(R-r). 

4. (a+6) 2 -c 2 = (a+6+c)(a+6-c). 

167. Incomplete Squares. Sometimes an expression comes 
under this type, but it is not stated directly as the difference 
of two squares. 

Thus, factor a 4 +a 2 6 2 +6 4 . 

This expression would be the square of a 2 -f-6 2 if the middle 
term were 2a 2 6 2 instead of a 2 b 2 . 

We will then add a 2 b 2 to complete the square and subtract 
it again to maintain the value of the expression. 

Thus, a 4 +a 2 6 2 +6 4 = (a 4 + 2a 2 6 2 +6 4 )-a 2 6 2 =(a 2 +6 2 ) 2 -(a6) 2 . 
= (a 2 + b 2 +ab)(a 2 +b 2 -ab). 
Examples: 

1. z 4 +x 2 + l = {x i +2x 2 + l)-x 2 . 

= 2 + l) 2 -x 2 . 

= {x 2 + l+x){x 2 + l-x). 

2. z 4 +9x 2 +25 = Or 4 +10z 2 + 25)-a; 2 . 

= 2 +5) 2 -z 2 . 

= (x 2 +5+x)(x 2 +5-x). 

3. 4x 4 + l = (4:r 4 +4.r 2 + l)-4:r 2 . 

= (2z 2 + l) 2 -(2z) 2 . 

= (2x 2 + 1 + 2x) (2x 2 + 1 - 2x) . 

Exercises CI. 

Factor: 

1. lQx 2 -25y 2 . 6. x i -y i . 

2. x 2 -9y 2 . 7. x s -y s . 

3. 25zy-16a 2 6 2 . 8. a 2 -6 2 -26c-c 2 . 

4. 9x 2 y 2 -4:p 2 q 2 . 9. (x+y) 2 -{a+b) 2 . 

5. a 2 -(6+c) 2 . 10. (x 2 +y 2 ) 2 -±x 2 y\ 



268 MATHEMATICS FOR TECHNICAL SCHOOLS 

11. x 2 -y 2 +2yz-z 2 . 20. x 4 +9x 2 + 81. 

12. l-a 2 -2ab-b 2 . 21. x 4 +4?/ 4 . 

13. a 16 -l. 22. x*-7x 4 + l. 

14. x 2 -2:n/ + ?/ 2 -a 2 -2a&-& 2 . 23. 4.r 4 -37xV +9*/ 4 . 

15. 7rl0 2 -7r7 2 . 24. x 4 +4x 2 +16. 

16. 7r7-5 2 -7r2-5 2 . 25. 9x 4 -IOxV+i/ 4 . 

17. 5a 2 -10a6+56 2 -20c 2 . 26. 4a: 4 -13x 2 2/ 2 +92/ 4 . 

18. 16-a 2 -6 2 +2a6. 27. x 4 +5x 2 2/ 2 +9?/ 4 . 

19. 4x 4 + llxV+9?/ 4 . 28. x 4 +x 2 -f25. 

168. Fourth Type. Divide x 3 -\-y 3 by x+y. 
x+y)x 3J ry z /x 2 — xy+y 2 





x 3 
x- 


x 3 -\-x 2 y 




— x 2 y+y 3 

— x 2 y — xy 2 


Divide 


xy 2 +y 3 
xy 2 +y 3 

-y 3 by x — y. 
-y)x 3 —y 3 /x 2 +xy+y 2 
x 3 — x 2 y 




x 2 y — y 3 
x 2 y — xy 2 




xy 2 — y 3 
xy 2 -y 3 



As a result of the above we may write : 

x 3 -\-y 3 = {x-\-y)(x 2 — xy-\-y 2 ) and x 3 — y 3 = {x—y){x 2 — xy-\-y 2 ). 

The above results might be stated as follows: 

The sum of the cubes of two quantities is divisible by the sum 
of the quantities, and the difference of the cubes of two quantities 
divisible by the difference of the quantities. The other factor 
consists of the sum of the squares of the quantities, minus their 
product, if the sum of two cubes, and plus their product if the 
difference of two cubes. 






RESOLUTION INTO FACTORS 2(59 

Examples: 

1. p 3 +q z = (p+q)(p 2 -pq+q 2 ). 

2. 8x 3 -27y 3 =(2xy-(Zyy = (2x-3y)(4:X 2 +Qxy+9y 2 ). 

3. 5a 3 -40 = 5(a 3 -8) = 5(a-2)(a 2 +2a+4). 

Exercises CII. 
Factor: 

1. 2/ 3 +27. 5. x 6 -64. 9. 2x 3 -f-250. 

2. a 3 -125. 6. a 3 -216. 10. x l2 -y 12 . 

3. x 6 + l. 7. 3-81x 3 . 11. (a+6) 3 -c 3 . 

4. a 6 -6 6 . 8. z 4 -27z. 12. (a+6) 3 -(o-6) 3 . 



CHAPTER XVIII. 
INDICES AND SURDS. 

169. Indices. In the introductory chapter in Algebra we 
inferred the laws with respect to indices from particular cases. 

Thus, (1) x 3 Xx 2 = x 3+2 = x 5 . (3) (f) 2 = x 3 Xx 3 =i 8 . 

(2) x 5 -Hx 2 = x 5-2 = x 3 . (4) (xy) 2 = xyXxy = x 2 y 2 . 

In the following discussion general proofs will be given for 
these laws and also their application when the indices are 
fractional, zero, or negative. 

Definition. If x is any number and m any positive integer 
x m means the product of m factors each equal to x. 

1 . To prove x m X x n = x m+n - 
By definition: 

x m Xx n = (xxxxx. ...torn factors) X (xxxxx to n factors). 

= xxxxx to (ra-f-n) factors. 

= x m+n ky definition. 

From the above it follows that: 

x m xx n xx v = x mJrn xx v = X m ^~ n ^~ V m 

x m 

2. To prove — =x m ~ n m>n. 

By definition: 

x m 

— = (xxxxx. ...torn factors) -j- (xxxxx ton factors). 

= xxxxx to (ra — n) factors. 

= x m ~ n by definition. 

3. To prove {x m ) n = x mn - 

{x m ) n = (x m ) X (x m ) X (x m ) to n factors. 

_ gjn+m+m .... to n terms k„ i 

= x mn - 

270 



INDICES AND SURDS 271 

4. To prove (xy) m = x m y m - 

(xy) m = (xy) X (xy) X (xy) .... to m factors. 

= (xxxxx. . to m factors) X (yxyxy. ..torn factors). 
= x m Xy m = x m y m . 
The above are known as the fundamental laws of indices. 

In assigning a value to x m , the definition requires that m 
be a positive integer, so that x*, x~ 3 , x° have as yet no meaning. 
However, one of the advantages of Algebra over Arithmetic 
is that it extends the principles of Arithmetic to negative 
numbers, so in harmony with this principle we will assume 
that the laws proved for positive integral indices holds for 
negative and fractional indices. 

(a) Meaning of x°. 

Since x m Xx n = x m+n for all values of m and n, if we replace m 
by 0, we have x° Xx n = x n+0 =x n . 

x n 
x n 

This relation was assumed when we said that log 1=0, 
for by the above 10 = 1, therefore by definition of logarithm, 
log 1=0. 

(6) Meaning of x~ n . 

Since x m Xx n = x m+n for all values of m and n, if we replace 
m by — n, we have: 

x~ n Xx n = x~ n+n = x° . 
But x° = l. 

x n 

From the above it follows that any factor may be transferred 
from the numerator to the denominator of an expression, or vice 
versa, by changing the sign of the index. 



272 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Exercises CIII. 

Express with positive indices: 

1. x~ 2 . 4^ 

2. p~ e . 2-*' 
1 „ ■ 5" 

a; -3 * 

2 
3 - 2 - 

6- 1 
3- 2 - 



3. 
4. 

5. 



10- ^xS-^A- 



25- l * 
a~ 4 " 



11. 3~ 2 xjx3 3 

1 
12 



-xa 3 x— . 
1 2 „-a* 



9. a- 6 x— x— - 



a z a 



(c) Meaning of x q , p and q being positive integers. 

Since x m xx n = x m+n for all values of m and n, if we replace 
both m and n by §, we have x i xx i = a; J+i = a; 1 = a\ 

Thus if x^ be multiplied by a;* we get the product x, or 
otherwise stated the square of x* = x. 

We have, however, previously represented the quantity 
whose square is x by y/x. 

.'. X i = y/x. 

Similarly x^xx^xx* = # i+i+ * = x, 
.'. x* = \/x (cube root of x). 
Generally xn = Va;(nth root of x). 
Again, since (x m ) n =x mn , 
then (x*) 4 =x 3 , 
.*. x$ = -\/x 3 . 



Example 1 
Example 2 
Example 3 



Similarly, (x«) q = x p , , 

.*. x"= 1/x p , p and q being positive integers. 
16* = ^16 = ^2* = 2. 
27* = (^27) 2 = 3 2 = 9. 
64* = (a$/64) 5 = 2 5 = 32. 



INDICES AND SURDS 273 



Exercises CIV. 

Write with positive indices: 



1. a 3 6~ 2 . 




6. 2x*X3aT 1 . 




ii. ,/— 

12. 2. 

4r _1 

13. _, . 

x * 

14. 7a-*X3a" 1 . 


9 a " 2 v a3 
*' b-**b*' 

a 3 
3. - 2 X a~*. 
or 

a" 2 6- 8 

*' c-^-6* 




7. 2a-i- 

8.4,. 
x J 

« 2a -2 
9. r. 

a - * 




_ 2x~ l 
5 ' 4jT»" 




io. JL. 

Vz 3 




a - * 

6a 


Ifa = l, 6 = 


2, n = 


= 3, find the value 


of: 




16. (ab) n . 




19. (a n b n )\ 




22. (a 3 6 3 )- n . 


17. <£)\ 




20. (a-'ft- 1 ) - ". 




23. (a" 4 6) n . 


18. (a*b-y. 




21. (a" 2 6 2 )- n . 






Find the value of: 






24 2X6 " 2 

J4 ' 3" 2 ' 




27 ' F* 




30. (A 2 *)" 1 . 


25. 16?. 
28 - 25"*- 




28. (If)"*. 




31. (W-) 1 . 




29. 16 1 " 5 . 




32. 36"i 


Show that: 










33. 12* = 2X3 


i 


34. 108 4 = 3X2' 


I 


35. 80* = 2X5*. 


Express as 


the root of an integer: 






36. &XS*. 




37. 3 s X 9*. 




38. 3* X 9* 4- 27*. 


39. Multiply 


x>+y h hy x h -y h . 40. 


Multiply z* +2/* by x — y. 


Solve: 










41. x h = 2. 


42. 


aT* = 4. 43. 


1 


= 4. 44. z* = 27. 



170. Surds. 

Definition. If the root of a number cannot be exactly 
determined, the root is called a surd. 



274 



MATHEMATICS FOR TECHNICAL SCHOOLS 




Fig. 129 



Thus, y/2 is a surd because we cannot find a number whose 

square is exactly equal to 2. 
We can find its value to 
a number of decimal places 
(1-4142), but this is only an 
approximate value. 

We can, by a geometrical 
process (Fig. 129), find a line 
which is the y/2 units in length. 
If we draw two lines at right 
angles to each other and each 
1 unit in length, then the 
hypotenuse of the right-angled 
triangle so formed would be 
the y/2 units in length. By 
continuing as in diagram, lines a/3, \/4, etc., may be found. 

171. Quadratic Surds. We are chiefly concerned with surds 

in which the square root is to be found. These are called 

t 
quadratic surds. 

Thus, \/2, a/3, V6, a/8 are quadratic surds. 

172. Surds other than Quadratic. These are indicated by 
the root symbol. 

Thus, ^6, ^9, AyiO, the first being called a surd of the 
third order, the second a surd of the fourth order, the third a 
surd of the fifth order. 

A surd is sometimes called an irrational quantity, and for 
the sake of distinction, quantities which are not surds, are 
called rational quantities. 

173. Like and Unlike Surds. When surds in their simplest 
form have the same surd factor they are called like surds, 
otherwise they are unlike surds. 



INDICES AND SURDS 275 

Thus, 2a/3~, 3V3", 5 a/3 are like surds, and 2 a/3, 3a/2^ 
5 a/6 are unlike surds. Just as we add and subtract like terms 
in Algebra, so we may add and subtract like surds. 

Thus, 2 a/3 +3 a/3 = 5 V3. 

5a/2-3a/2 = 2a/2. 5 a/6 +2 a/6 -3 a/6 = 4 a/6. 

174. Multiplication of Surds. Since a/3 represents a quantity 
whose square is 3, .*. a/3 X a/3 =3. 

Again, since (a/3X a/2) 2 = a/3X a/3X a/2 X a/2. 

= 3X2 = 6. 

.'. a/3 X a/2 = a/6. 

Similarly, a/3 X a/5 = V lT. 

Generally, a/« X a/& — \/a&- 
In the above the surds multiplied together are of the same 
order. If, however, we wished to multiply a/2 and a/3, it 
would first be necessary to change them to surds of the same 
order. 

By the previous section on indices: 

V2 = 2* = 2?, a/3 = 3* = 3*. 

.*. V2X^3 = 22X3* = Ay2 3 XAy3 2 . 

= a/8 X a/9 = a/72. 

Exercises CV. 

Express as surds of the same lowest order: 

1. a/3, a/4, a/6.^ 3. a/2,V8, a/4. 

2. a/5\ \/ll, a/18, 4. a/* a/3", V6. 

Find the product of: 

5. a/2, a/3. 8. a/5, a/6. 

6. a/3, a/5, a/2. 9. a/2, a/3, a/4. 

7. Vf, Vf, Vf 10. V3, V5, a/6. 



276 MATHEMATICS FOR TECHNICAL SCHOOLS 

175. Mixed and Entire Surds. When a surd quantity is 
the product of a rational quantity and a surd, it is called 
a mixed surd. If there is no rational factor it is called an 
entire surd. 

Thus, 6\/3 is a mixed surd, and \/7 is an entire surd. 

The expressing of a mixed surd as an entire surd would 
be of little value practically, but the reverse process is of 
frequent application. 



Thus, V27 = V9X3=3 v / 3. 



Again, V72= V36X2 = 6\/2. 

Exercises CVI. 
Express as a single surd: 

1. 2 V63 + 5 V28 - V7. 3. V72 + V98 - V128 + V32 + V50- 

2. 10V44-4V99. 4. V45-V20 + V80. 
Find the value correct to two places of decimals: 



5. V288. 11. V36-V72 + V90. 

6. V147- 12. 4V63+5V7-8V28. 

7. V250. 13. 2V363-5v / 2434-V'192. 

8. 3V150. 14. 5V24-2V54-V6. 

9. 5V245. 15. 4V128+4V75-5V162. 
10. 4V63". 

Express in simplest form: 



16. ^256. 17. ^432. 18. ^3125. 19. ^/-2187. 

Find the value to two decimal places: 

20. 2V14XV2T. 24. 2Vl4X3\/28. 

21. 3V8XV128. 25. 2V15X3V5. 

22. V50XV75. 26. 8V12X3V2T 

23. 3V6X4V2. 



INDICES AND SURDS 277 

176. Division of Surds. 

Since y/x X Vy = Vxy, 

Si mil arly , y/x -i- y/y = % -, 
\y 

and 2V30-4-3V6 = 3\'y = ^V5. 

Example: — Find the numerical value of— j=- (tan 30°). 

We might find the square root of 3 and perform the division. 
This, however, would not be the best method, 

For-UJ_ x V3 = V3 = L7321 = . 5774 . 

V3 \/3 V3 3 _3 

1 V3 

Here we changed ~~k into J ~- by multiplying both numera- 
v o o 

tor and denominator by \/3. 

This operation of making the denominator a rational 

quantity is called rationalizing the denominator. 

1 
Example: — Find the value of — ;= (sin 45°). 

-UJLx^-^-i^-Ton. 

V2 V2 x/2 2 2 

Example: — To rationalize the denominator of an expression 

of the form ^-^ 

2-V2. 

Here we wish to convert = into an equivalent expres- 

2 — s/2 
sion but with a rational denominator. 

Since the product of the sum and difference of two quanti- 
ties is equal to the difference of their squares, then (2 — \/2) 
(2 + V2) =4-2 = 2. 

. l + V2 x 2 + v / 2 _ (l + V2)(2 + V2) ^ 4+3V2 
2-V2 2 + V2 2 2 



278 MATHEMATICS FOR TECHNICAL SCHOOLS 

The expression 2 + \/2 is known as the conjugate expression 
to 2 — \/2. If the denominator of the fraction had been 2 + y/2 
we would then have multiplied by 2— V2. 

Exercises CVII. 
Calculate the value of the following to 3 places of decimals: 

1. *L 8. -U 15. ***** 

V3 V500 7V2-3 

2 4 2 

2 -V3* 9 'VW 16 - 3 "v6- 

„ 12V2 ,n /25 - 

3 --^' 10 'V252- 17. (V3-V2) 2 - 

4 A 11. -i^ 18 . V3-1 
V5 2 "V2 V2-1 

5 4= 12- -=-5— • 19 3 ^1 1 

V24* V5 + V2 " 3V2-1* 

48 13 5+2V6 . 4V7+3V2 



V6^_ 6-2V6 V3-V2 

256 14 j 

1575* ' V5-1* 



V 



CHAPTER XIX. 

QUADRATIC EQUATIONS. 

177. Quadratic Equations. The following problems will lead 
to equations which differ somewhat from those previously 
solved. 

Problem 1: 

The area of a square is 64 sq. in. What is the length of 
a side? 

If x represents a side of the square, then x 2 = 64. 

In the equations previously met the unknown x occurred 
to only one power, and that the first. Here, however, the 
unknown occurs to the second power. When an equation 
contains the square of the unknown quantity, but no higher 
power, it is called a quadratic equation. 

In the equation x 2 = 64 we have the simplest form of the 
quadratic equation: 

If x 2 = 64. 
then x = ±S. 

We have here two values of x, i.e., + 8 and— 8, which will 
satisfy the equation. If we regard -f- and — as opposite 
directions in the same straight line, the minus value has no 
significance in determining the side of the square. 

Problem 2: 

The length of a number-plate on a machine is 6" more than 
its width. If its area is 72 sq. in., find its dimensions. 
If x = No. of in. in width, 
then x + 6 = No. of in. in length, 
then x (x + 6) =72. 
or x 2 + 6.r-72 = 0. 
279 



280 MATHEMATICS FOR TECHNICAL SCHOOLS 

By our previous principles in factoring x 2 -\-6x — 72 = (ar-f-12) 
(x-6). 

Now, if z 2 +6z-72 = 0, then (a? + 12)(a:-6)=0. 

In order that the product of these two factors may be equal 
to zero, it is necessary that one factor should be equal to 
zero. 

Thus the equation will be satisfied if x + 12 = or x — 6 = 
or if x = —12 or 6. 

As + 6 is the only admissible value, therefore the width 
= 6" and the length 6+6 = 12". The equation z 2 +6.r-72 = 
is known as a complete quadratic equation, containing as it 
does both the square and the first power of the unknown 
quantity. 

Exercises CVIII. 

Solve the following equations and verify: 



1. 


x 2 = 49. 




11. 


x 2 -2 = x. 


2. 


x 2 +3x = 0. 




12. 


4x =45 — x 2 . 


3. 


x 2 = 7x. 




13. 


5x 2 -12z+4 = 0. 


4. 


z 2 -4 = 0. 




14. 


3z 2 + 14x-15 = 0. 


5. 


6x 2 = 54. 




15. 


20x 2 +4Lr+20 = 0. 


6. 


z 2 -3 = l. 




16. 


5+9x-2a; 2 = 0. 


7. 


z 2 -10z+21 = 


0. 


17. 


18a: 2 -9x -2 = 0. 


8. 


z 2 -14z+48 = 


0. 


18. 


13z 2 +4Lt + 6 = 0. 


9. 


a; 2_ a ._20 = 0. 




19. 


12a: 2 -x- 20 = 0. 


10. 


z 2 +10 = llx. 




20. 


6x 2 -x-2 = 0. 



178. Solving by Completing Squares. In connection with 
the squaring of a binomial we recall that (a + 6) 2 = a 2 + fe 2 + 2a6, 
or that the square of a binomial equals the square of each 
term, plus twice their product. If then we have x- +6.r and 
we wish to add a sufficient quantity to make a complete square, 
we could reason as follows: # 2 is the square of x, Qx is twice 
the product of x and 3, therefore it is necessary to add 3 2 or 9. 

.*. a: 2 +6x+9 is a complete square = (x+3) 2 . 



QUADRATIC EQUATIONS 281 

Similarly, to x z — 8x or x 2 -2XxX4 we must add 4 2 or 
16, giving x 2 — 8x + 16 = (x — 4) 2 . 

Again, to x 2 +Qx, or x 2 +2X^X| we must add (f) 2 or ^, 
giving a; 2 +9a;+-^- = (a;+f) 2 . 

An analysis of the three cases above would lead us to infer 
that we completed the square in each case by adding the square 
of half the coefficient of x. 

This method is necessary where the quadratic equation 
cannot readily be resolved into factors. 
Thus, Example 1:— Solve x 2 - 6a; - 13 = 0. 

or, a: 2 — 6a; =13. 
Completing the square on the left-hand side we have: 
a; 2 -6a;+9 = 13+9. 
or, (x-3) 2 = 22. 
Extracting square root, x — 3 = ±22. 

.*. x = 3 + V22 or 3-V22. 
= 7-69 or -1-69. 
Example 2:— Solve -3a; 2 +8a; + 12 = 0. 

In the examples above on completing the square, we observe 
that the coefficient of x 2 in each case is unity and further that 
it is positive. Before attempting then to solve this equation 
we must make these two changes. 

-3x 2 +8x + 12 = 0. 
= 3a; 2 -8a; -12 = 0. 
= x 2 — fa;— 4 — 0. 
Complete the square, giving: 

* 2 -t*+a) 2 =4+(!) 2 . 

or, (a;-!) 2 ^-^ 
or, *-$'-* ±V¥-'< 

or, x = $±V-^. 

= l + V-V-or|-V¥-. 
= 3-74 or -1-07. 



282 MATHEMATICS FOR TECHNICAL SCHOOLS 

Exercises CIX. 
Solve by completing the square: 

1. 5x 2 + 14a; -55 = 0. _J_ 1 _ 6 

2. 9x 2 -143-6x = 0. l+a;~3^-a; = 35' 

3. 19x = 15-8a; 2 . 5 4 3 

15. 



4. 6x 2 -9x-15 = 0. x-2 x~x+6" 

5. 5x 2 +llx-12 = 0. _^_ 5 3 

6. 2x 2 + 7-9x = 0. lb ' x -l~a;-f 2~x" 

_*±3_2x-l 
2x-l x-3 

18. 21x 2 -2ax-3a 2 = 0. 

19. 12x 2 +23fcx + 10fc 2 = 0. 
1 5 2 



7. 


5x 2 -15x + ll = 


8. 


x 2 -7x+5 = 0. 


9. 


x 2 +ll = 7x. 


10. 


4x 2 = ^x+3. 


11. 


/v.2 O — 2 3/h 


12. 


a; — 1 


13. 


5a; — 1 _ 3x 
x + 1 2 ' 



20. 



2x — 5a 2x — a 



179. The General Quadratic Equation. From the preceding 
examples it is apparent that every quadratic equation can be 

reduced to the form 

ax 2 -f&x+c = 0, 
where a, b, c may have any numerical values whatever. If 
then we would solve this general quadratic equation we could 
use the result as a formula to solve particular cases and con- 
sequently save the labour entailed. 

ax 2 + 6x+c = 0. 
Transposing, ax 2 +bx= —c. 

dividing by a, x 2 -\ — x = — . 

Completing the square by adding to each side the square 

of half the coefficient of x, i.e., («-) ' 

bx . / b \2 b 2 



, . OX ( \2 0' C 

giving * 2 + a +( 2a ) = 4a2 " a 



or, 



(■*=)'- 



2a> 4a 2 



QUADRATIC EQUATIONS 283 



Extracting the square root 



Hence, x = 



b ±Vb 2 -4ac 



2a 2a 

±V& 2 -4ac 



2a 

We might here restate the steps required in solving a quad- 
ratic of the above form. 

(1) Simplify the equation so that the terms in x 2 and x are 
on one side of the equation, and the term without x on the other. 

(2) Make the coefficient of x 2 unity and positive by dividing 
throughout by the coefficient of x 2 . 

(3) Add to each side the square of half the coefficient of x 
thus completing the square. 

(4) Take the square root of each side. 

(5) Solve the resulting simple equations. 

Example 1: — Solve x 2 — 5x — 3 = 0. 

Here, a = l, b= — 5, c= — 3. 

5±V(-5) 2 -4XlX(-3) 
.*. x = ~ • 



= 5±V25 + 12 
2 

_ 5±V37 _ 5+6-08 
" 2 2 

= 5-54 or— -54. 

Example 2 :— Sol ve x 2 - 2x + 5 = 0. 

Here, a = l, 6= —2, c = 5. 



2d=V(-2)-4XlX5 

X = 7i • 



2±V4-20 
2 

2±V-16 



284 MATHEMATICS FOR TECHNICAL SCHOOLS 

In the preceding result the numerical value of the roots 
cannot be found, as there is no number whose square is 
negative. 



Such a quantity as y/ — 16 is called an imaginary quantity, 
and the roots are said to be imaginary. This is equivalent 
to saying that there is no real number which will satisfy the 
equation x 2 — 2x + 5 = 0. 

180. There are some Equations that are not really 
quadratics but may be solved by the methods of this 
chapter. 

Example 1: — Solve x 4 — 5x 2 +4 = 0. 

Factoring (x 2 — 4) (x 2 — 1) = 0. 

.\ x 2 = 4 or x 2 = 1. 

.'. x = ±2 and x = ±l. 

72 

Example 2 :— Solve x 2 - x + -j— - = 18. 

Write y for x 2 — x, then we have: 

72 
Hh-=18. 

or, y 2 -18y+72 = 0. 

Factoring, (y-12)(y-Q) =0. 

giving y = 12 or 6. 

.". x 2 — x = 12 or 6. 

If x 2 -x = 12 then a; 2 -x- 12 = 0. 

then(x-4)(x + 3)=0. 
giving x = 4, or —3. 
If x 2 — x = 6, then x 2 — x — 6 = 0. 

then (x-3)(x+2)=0. 
giving x = 3 or —2. 



QUADRATIC EQUATIONS 285 

Exercises CX. 

Solve the following examples: 

1. 3x 2 -17x + 10 = 0. 8. x 4 -13x 2 +36 = 0. 

3. 2 (x 2 + l)-5x = 0. y " 15 + 

4. 25x 2 -7z-86 = 0. 20 _ c 

5. 7x 2 +32a;-15 = 0. 1U ' x ~^ 6X x 2 +3x " 

6. (2x-l) 2 = 25. 11. (x 2 4-2) 2 + 198 = 29(x 2 +2). 

7. 10z 2 = 13x + 9. 12. a: 6 -19a: 3 -216 = 0. 

13. A rectangular name-plate for a machine is to be \\" 
longer than it is wide and to have an area of 10 sq. in. What 
will be its dimensions? 

14. Three holes are to be drilled so that they will lie at the 
three corners of a triangle ABC, right angled at B. The 
distance from A to G is to be 10" and the distance from 
B to C is to be 2" more than from A to B. Find AB 
and BC. 

15. The sides AB, BC, CA of a triangle measure 13, 14, 
15 respectively. From A a perpendicular AD is drawn to 
BC. If BD measures x, express the length of AD in two 
ways. Equate the results and find x. 

16. The owner of a rectangular lot 15 rods by 5 rods, 
wishes to double the size of the lot by increasing the length 
and the width by the same amount. What should be the 
increase? 

17. A straight line is 10" long. Divide it into two parts 
so that the rectangle contained by the whole line and one of 
the parts is equal to the square on the other part. 

18. S = \gt 2 is the law governing a body falling from 
rest, s = space, g = acceleration due to gravity (32 ft.), 
< = time in seconds. How long will it take a stone to fall 
from the top of the City Hall tower, Toronto, if it be 305 
ft. high? 

19. S = ut-\-\gt 2 is the law for a falling body when it has 
an initial velocity, u representing this initial velocity. If 
a stone be thrown with an initial velocity of 8 ft. per sec. 
from the top of the Eiffel tower, 984 ft. high, in what time 
will it reach the ground? 



28G MATHEMATICS FOR TECHNICAL SCHOOLS 

181. Simultaneous Quadratic Equations. The following 
problems will lead to simultaneous equations where one at least 
is of higher degree than the first. 

Problem : 

The perimeter of a rectangle is 18'', and its area is 20 sq. 

in.; find its length and breadth. 

If x represent the length and y the breadth, then: 

2z+2y = 18. 

or, x + y = 9 (a). 

also, xy = 20 (6). 

Solution — 1st method. 

from (a), y = 9— x. 

Substitute in (6), x(9-x)=20. 

or, 9z-z 2 = 20. 

or, z 2 -9x+20 = 0. 

or, (x — 5)(x — 4) =0. 

x = 5 or 4. 

20 . 20 _ 
.'. 2/ = -=4or T = 5. 

x = 5 or 4. 
2/ = 4 or 5. 
Solution — 2nd method. 

(a) 2 = x 2 +2xy+y 2 = 81. 
4x(6) = 4xy = 80. 
Subtracting, x 2 — 2xy+y 2 = 1. 
.*. {x-y) 2 =\. 
or, x— y = ±l. 
x+y=9 x+y=9 

x—y = l x—y=—l 



2x =10, x = 5. 2x =8, z = 4. 

2y = 8, y = 4. 2y =10, x = 5. 

.'. x = 5 or 4, ?/ = 4 or 5. 



QUADRATIC EQUATIONS 287 








Exercises CXI. 






Solve the following equations: 






1. x+y = 28, 


4. x+y = 84, 


7. 


x 2 +i/ 2 = 178, 


a;?/ = 187. 


xy = 92S. 




a; +2/ = 16. 


2. x-y = b, 


5. x 2 +y 2 = n, 


8. 


x 2 +y 2 = 185, 


xy = \2Q. 


xy = 35. 




x -y =3. 


3. x-y = 8, 
xy = 51S. 


6. x 2 +2/ 2 = 89, 
xy = 40. 


9. 


M-3, 



10. Divide a straight line 7" long into two parts so that the 
rectangle contained by the parts may be equal to 12 sq. in. 

11. Divide a straight line 12" long into two parts so that 
the sum of the squares on the parts may be equal to 74 
sq. in. 

12. The hypotenuse of a right-angled triangle is 25", and 
the perimeter is 56", find the sides. 



CHAPTER XX. 

VARIATION. 

182. Quantities are often related to each other in such a 
way that any change in one quantity produces a corres- 
ponding change in the other. 

For example, consider a train travelling with a uniform 
speed. If in one hour the train travels 30 miles, then in 
two hours it will travel 60 miles, and so on. 

We may state this by saying that the distance travelled is 
proportional to the time, or that the distance varies directly 
as the time. 

If we represent distance by d and time by t, the relation 
may be expressed by d varies as t. 

The symbol oc represents "varies as", therefore we have 
d oc t. 

If we let d 1} d 2 , d 3 be successive values of d and 

h» t*> 'a • • • corresponding successive values of t, then, 
we have: 





d t , d. 




t-=- or d= T L t. 

d i tx <i 


also, 


d t , d„ A 
j- = — or d = - ? t. 
u 2 *2 *2 


also, 


d t , d„ J 
j-= — or d=—*t, etc. 
a 3 t 3 t 3 



AAA 

Let us consider the expressions y, -~t / in the light 

t t t 2 t 3 

of the above illustration. 



288 



VARIATION 289 

If d, = 60, < x ~2, then^i =^° = 30. 

Ifd 2 =90, t 2 =S, then^- 2 =^ = 30. 

If d 3 = 120, * 3 =4, then ^=^ = 30. 

From the above we see that the ratios — L , -^, -^ are each 

t 1 t 2 t 3 

equal to 30, therefore we infer that doct becomes d = 30t 

or generally d = Constant x t or d = kt (k being a constant). 

Example 1: 

The circumference of a circle varies directly as its diameter. 
A circle 7" in diameter has a circumference of 22", find the 
circumference of a circle of 64" diameter. 
From the above C = kd, 
then, 22 = £7, 

or, fc = y, 

,.c=fz). 

Substituting for D the value 64, 

22 
C = ^X64 = 20M4". 

In the above we observe that the first set of conditions 
enable us to find the constant k. The equation is then one 
between C and D, and from any value of one of these we can 
find the other. 

Example 2: 

The areas of circles vary directly as the squares of their 
radii. If a circle with a radius of 7" has an area of 154 sq. 
in., find the radius of a circle with an area of 1386 sq. in. 

From the above Acer 2 . .'. A = kr 2 , then 154 = A; 7 2 , giving 

kJ* 
* 7* 



290 MATHEMATICS FOR TECHNICAL SCHOOLS 

Then substituting for A the value 1386 we have: 

22 
1386= r 2 . 

.\ r = 21. 

Example 3: 

The volume of a gas varies inversely as the height of the 
mercury in the barometer. If the volume is 22 cu. in. when 
the barometer registers 30", what is the volume when the 
barometer registers 32"? 

Here we have a case of varying "inversely." This means 
that an increase in one quantity gives a proportionate decrease 
in the other. Hence, when one quantity varies inversely as 
another it varies as the reciprocal of the other. 

In the above V oc^j or V=kjj. 

£1 tl 

k 
From the conditions given, 22 = — , giving A- = 660, 

Ox) 

then, F= o ° o u = 20-6 cu. in. 

Exercises CXII. 

1. The strength of a beam varies directly as the square of 
its thickness. 

A beam of given length and width and 6" thick carries a 
maximum load of 5 tons. What load will a beam of the 
same width and length, but 12" in thickness carry? 

2. The weight of a substance varies directly as its volume. 
A steel bar containing 100 cu. in. weighs 28-3 lb. What is 
the weight of a bar of the same material containing 642 
cu. in.? 

3. The velocity of a falling body varies directly as the time 
during which it is falling. When a body falls from rest, its 
velocity at the end of 1 sec. is approximately 32 ft. per second. 
Compute its velocity at the end of 15 seconds. 

4. The velocity of the rim of a pulley varies directly as its 
diameter. A 12" pulley has a rim velocity, at a certain 
moment, of 160' per min. What is the rim velocity, at the 
same moment, of a 9£" pulley which is keyed to the same shaft? 



VARIATION 291 

5. The deflection of a beam under a given load varies 
inversely as the square of the thickness. If a given beam 
carrying a certain load is 5" in thickness and has a deflection 
(due to the load) of 2", what deflection will be produced by 
the same load if the thickness be 7§"? 

6. The weight of a body varies inversely as the square of 
its distance from the centre of the earth. If a substance 
weighs 10 lb. at sea level (3960 miles from the centre), compute 
its weight on the top of a mountain 29,000 ft. above sea level. 

7. The areas of circles are to one another as the squares of 
their diameters. If a circle with a diameter of 14" has an 
area of 154 sq. in., find the diameter of a circle with an area 
of 320 sq. in. 

8. The pressure per sq. in. on a hydraulic ram varies inver- 
sely as the square of the diameter, if the total load on the ram 
is constant. If a load supported by a hydraulic ram of 8" 
diameter gives a pressure per sq. in. of 40 lb., find the pressure 
per sq. in. on a 3|" ram which supports an equal load. 

9. The horse-power of the engines of a ship varies directly 
as the cube of the speed. If the horse-power is 1800 at a 
speed of 10 knots, what is the power when the speed is 23-5 
knots? 

10. The volumes of spheres vary directly as the cubes of 
their diameters. If a sphere with a diameter of 6" has a 
volume of 1134- cu - m -> find the diameter of a sphere whose 
volume is 616 cu. in. 

11. The number of rivets required for a boiler seam varies 
inversely as the pitch (the distance between rivet centres). 

If 35 rivets are required when the pitch is 2| ", determine the 
pitch when 40 rivets are required for a boiler of the same size. 

12. The resistance of a wire varies inversely as the square 
of its diameter. 

The resistance of a coil of copper wire \" in diameter was 
3 ohms. What is the diameter of a wire of the same length 
with a resistance of 2-3 ohms? 

183. Problems involving more than Two Variables. If we 

take two rectangles of the same width, it is readily seen 
that their areas vary as their lengths. If again we take two 
rectangles of different widths but the same length, it is 
further agreed that their areas vary as their widths. 



292 



MATHEMATICS FOR TECHNICAL SCHOOLS 



We now wish to consider the variation in the area when 
both the length and width vary. 



i 



I— I 



A, 



W 




Fig. 130 

In Figure 130 above the rectangles (1) and (2) have the 
same width but different lengths, while the rectangles (2) 
and (3) have the same length but different widths. 

If A, A 1} and A 2 represent the respective areas, then with 
the above notation: 

A Iw _ I 
A 1 ~l 1 w l t 
A x l x w w 
A 2 l 1 w l w x 



also, 



(a), 



, '\ w/1 v A A. I w 
(«)X(6) -j-x-ji-j-*-. 

A Iw . A o , 

or, - -.- = : ■ or A = -, — — Iw. 

A 2 l 1 w x l 1 w l 

Since, A a =l 1 w l .'. j—* =1 (constant). 
l l w l 

.'. A = Constant X Iw. 

.". A oc Iw. 

From this we state that the area of a rectangle varies as 
the product of its length and width, when both the length 
and width vary. 

Further we know that triangles of the same altitude are 
to one another as their bases, and also that triangles of equal 
bases are to one another as their altitudes. Hence we might, 
as in the case of the rectangle, prove that the area of a triangle 
varies as the product of the base and altitude when both base 
and altitude vary. 



VARIATION 293 

Again, the volumes of cylinders of the same height are to 
one another as their bases, and also the volumes of cylinders 
with equal bases are to one another as their heights. Hence 
it might be proved that the volume of a cylinder varies as the 
product of the base and height, when both base and height 
vary. 

From these illustrations we infer the general theorem: — 
If A varies as B when C is constant, and A varies as C when 
B is constant, then A varies as BC when both B and C vary. 

Definition — One quantity is said to vary jointly as a number 
of others when it varies directly as their product. 

Example 1: 

The volume of a cone varies jointly as its altitude and the 
area of its base. The volume is 392-7 cu. in. when the alti- 
tude is 15" and the diameter of the base 10". Find th.e 
diameter when the altitude is 22" and the volume 436 
cu. in. 

Here Foe AB or V = kAB. 

Substituting the first conditions: 

392-7 = * 15X-7854X10 2 . 

giving fe- 15x . 7864xl0 r 
From the second conditions: 

436= 1 5 X 3 7854 7 XW X22B - 

436X15X-7854X10 2 
22X392-7 

If d be required diameter, 

436X15X-7854X10 2 



then -7854 d 2 = 
.'. d* = 



22X392-7 
436X15X10 2 



22X392-7 
d = &-7". 



294 MATHEMATICS FOR TECHNICAL SCHOOLS 

Example 2: 

The volume of a gas varies inversely as the pressure and 
directly as the absolute temperature (the absolute tempera- 
ture is obtained by adding 273 to the temperature on the 
Centigrade scale). 

If a quantity of nitrogen under 900mm. pressure at 20° C. 
occupies a volume of 300cc, what volume will it occupy at 
100° C. under 600mm. pressure? 

T T 

Here, V ex p or V = k p. 

From first conditions: 

onn . 293 . . . 300X900 
300 = fc 9 ^, giving Jc = -293-- 

From second conditions: 

T , 300X900 373 __« _ 
F = --293-" X 600 = 572 - 87cC - 

Exercises CXIII. 

1. The area of a triangle varies jointly as its base and alti- 
tude. The area of a triangle whose base is 19' and whose 
altitude is 10' is 95 sq. ft. Find the altitude when the base 
is 22-5' and the area 134 sq. ft. 

2. The volume of a pyramid varies jointly as its height 
and the area of its base. When the height is 18' and th(> base 
a square 8' to the side, the volume is 384 cu. ft. What is 
the side of the base if a pyramid of the same form, 10' high, 
has a volume of 432 cu. ft.? 

3. The pressure of the wind perpendicular to a plane surface 
varies jointly as the area of the surface and the square of the 
velocity of the wind. Under a velocity of 16 miles per hour 
the pressure on 1 sq. ft. is 1 lb., what is the velocity when 
the pressure on 3 sq. yd. is 68 lb.? 

4. The amount of illumination received by a body varies 
directly as the intensity of the light and inversely as the square 
of the distance from the light. 



VARIATION 295 

From a light of 14 candle-power, the illumination is 5 at 
a distance of 8'. Find the illumination at a distance of 10' 
from a light of 40 candle-power. 

5. The intensity of a magnetic field varies directly as the 
number of vibrations and inversely as the square of the dis- 
tance of the magnet. When the distance is -5", and the 
number of vibrations 15, the intensity is • 14. Find the 
intensity when the distance is -075", and the number of 
vibrations 90. 

6. The heat developed in a conductor varies jointly as the 
resistance of the conductor, the time the current flows, and 
the square of the current. In 2 minutes a current of 4 amperes 
developed 1400 units of heat in a wire having a resistance of 
11 ohms. Find the resistance of a wire of the same size in 
which 20,000 units of heat were developed by a current of 
6-84 amperes in 3£ minutes. 

7. The stiffness of a rectangular beam varies jointly as the 
breadth and as the cube of the depth. Show that two beams 
of the same material of breadth and depth (1) 2", 3", (2) 1" 
3-78", are of nearly the same stiffness. 

8. If the attraction between two masses varies directly as 
their product and inversely as the square of the distance 
between their centres, what would 1 lb. weigh on the surface 
of a planet of the same density as the earth, but 1 • 5 times the 
diameter? 

9. The square of the time which a body takes to slide down 
an incline varies as the square of the length and inversely as 
the height. If the time taken is 1 sec. when the height is 
4' and length 8', what must be the height of a plane 3' long 
so that the body may slide down in \ sec? 

10. The volume of a cylinder varies jointly as its 
base and height. Of two cylinders volume of 1st: volume 
of 2nd = 11: 8, and height of 1st: height of 2nd = 3:4. 
If the base of the first is 16-5 sq. ft., find the base of the 
second. 

11. The volume of a gas varies inversely as the pressure 
and directly as the absolute temperature. 

What would be the volume at 0°C. and 760mm. pressure 
of a mass of oxygen whose volume is 60cc. at 40C, under 
a pressure of 750mm. of mercury? 



296 MATHEMATICS FOR TECHNICAL SCHOOLS 

12. At what temperature must a gas be so that its volume 
will be 15 litres when the pressure is 800mm., if its volume 
is 175 litres when its temperature is 100°C, and the pressure 
700mm.? 

13. The electrical resistance of a wire varies directly as 
the length and inversely as the square of the diameter of the 
wire. Its weight varies jointly as the length and the square 
of the diameter. 

If a pound of wire of diameter -06" has a resistance of 
•25 ohms, what is the resistance of a pound of wire of the 
same material, the diameter being -01"? 



CHAPTER XXL 
GEOMETRICAL PROGRESSION. 

184. Amount. If I deposit $100 in a savings bank which 
pays interest annually at 4%, I will be entitled to $4 interest 
at the end of the first year. If I choose to leave this interest 
on deposit my bank account would then be $104. This sum, 
representing the principal plus the interest, is said to be the 
amount of $100 in one year. 

Consider the following examples. 

Example 1: 

Find the amount of $100 in 3 years at 6% per annum, com- 
pounded yearly. 

The interest at the end of the first year = -— of $100. 
The sum itself = j52 f $100. 

.'. the amount at the end of the first year= -^- of $100. 

= $100(1-06). 
The interest at the end of the second year = -^r of $100(1 • 06). 
.'. the amount at the end of the second year 

= j^| of $100(1-06). 

= $100(1 -06) 2 . 
The interest at the end of the third year = — — of $100(1- 06) 2 . 

.-. the amount at the end of the third year = ---pi $100(1 • 06) 2 

= $100(1 -06) 3 

= $119-10. 
297 



298 MATHEMATICS FOR TECHNICAL SCHOOLS 

Example 2: 

A man saves $200 a year for 4 years. If each year, he invests 
it at 6% per annum, what are his accumulated savings 4 years 
from the date of his first investment? 

The amount of the first $200 = $200(1 -06) 4 . 
The amount of the second $200 = $200(1- 06) 3 . 
The amount of the third $200 = $200(1 -06) 2 . 
The amount of the fourth $200 = 8200(1-06). 
Accumulated savings = $200(1 • 06) 4 + $200(1 -06) 3 
+$200(1 -06) 2 +$200(l -06), 
= $200(1- 06 + 1 -12360 + 1 -19102 + 1 -26248). 
= $200X4-63710. 
= $927-42. 
If in the preceding example the time had been 15 years, 
instead of 4 years, it is apparent that considerable work would 
be involved. The following discussion will develop a formula 
for shortening the work. 

Consider the Series: 

a-\-ar-\-ar 2 -\-ar 3 ar n ~ l . 

It is apparent that (1) each term is obtained from the pre- 
ceding by multiplying by r, called the common ratio, (2) in 
the second term r is raised to the first power, in the third term 
to the second power, .*. in the nth term to the (n — l) th power. 
(3) there are n terms in the series. 

Such a series is called a Geometrical Series and the terms in 
the series are said to be in Geometrical Progression. 

Let S = a + ar+ar 2 ar n - 2 + ar n ~ l . 

.'. rS= ar-\-ar 2 ar n ~ 1 -\-ar n . 

.-. rS-S = ar n -a. 

.-. S (r-l)=a (r n -l). 

" S= r-1- 

We here observe that in the above formula a is the first term, 
r the common ratio, and n the number of terms. 



GEOMETRICAL PROGRESSION 299 

Returning to our previous difficulty in finding the accumu- 
lated savings at the end of 15 years, we have as in Example 2. 

Accumulated Savings 

= $200(1 -06) 15 + $200(1 -06) u . . . .$200(1-06) 

= $200(1- 06 + (l-06) 2 ....(l-06) 15 )}. 

The series within the brackets is evidently a geometrical 
series, the first term being 1-06, the common ratio 1-06, 
and the number of terms 15. 

. « , n J(l-06) 15 -l\ n J 2-39656-l \. 

..Sum = 1.06(- 106 _ r ) = l-06( 1Q6 _ 1 ^ 

. . . , a $200X1 06 XI -39656 

.". Accumulated Savings = ^ 

• Ob 

= $4934.88. 

Note. — The value of (1-06) 15 may be obtained from interest 
tables or by the use of logarithms. 

185. Present Worth. If a person owes me $106 a year 
from now, and money is worth 6%, I might as well accept 
$100 now. 

The $100 is here called the Present Worth of $106 and we 
may, therefore, define the present worth of a future payment 
as the sum which will at the given rate, amount to the pay- 
ment when due. 

Example: 

A man wills his son $1000 a year for 10 years. What is 
this legacy worth now, if money is worth 5% per annum? 

The amount of $1000 one year hence = $1000(1 -05). 

.•.81000(1-05) due one year hence has for present worth 
SI 000. 

$1000 
.*. $1000 due one year hence has for present worth ' . 

1 -05 

The amount of $1000 two years hence = $1000(1 -05) 2 . 



300 MATHEMATICS FOR TECHNICAL SCHOOLS 

.". 81000(1 • 05) 2 due two years hence has for present worth 

$1000. 

$ 1 000 
.'. $1000 due two years hence has for present worth^ — -V,. 

(1 -05) 2 

Similarly $1000 due three years hence has for present worth 

$1000 



(1-05) 3 ' 
.*. Present Worth of all the payments equals 
$1000 $1000 , $1000 , $1000 



1-05 _r (l-05) 2_r (l-05) 3 ^(l-05) 10 ' 

$1000 fl + (l-05) + (l-05) 2 +(l-05) 9, 

$1000 

"(1-05) 10 

$1000^ 1-62889-1 $1000 -62889 

\ io X i . ns' _ i ""n.n^MoX .n* -$7721-54. 



Pfi^}] 



(1-05) 10/N 1-05-1 ~(1-05) 10/ ^ -05 

Exercises CXIV. 

1. Find the amount of $450 if left on deposit in a bank for 
3 years, if interest at 3% per annum compounded half-yearly 
be allowed? 

2. What sum of money loaned at 5% per annum will in 7 
years yield $407-10 interest? 

3. What sum will amount to $1986-86 in 17| years at 4% 
per annum? 

4. A person deposits $100 in a savings bank on January 
1st, 1911, and the same sum each year until January 1st, 1921. 
If banks pay 3% per annum, compounded half-yearly, what 
sum stands to his credit just after making the deposit on 
January 1st, 1921? 

5. A person holds $6000 in bonds paying 5% per annum. 
He dies leaving the income for the first 10 years to his son. 
If money is worth 6% per annum, what is the present worth 
of the legacy? 

6. A mortgage for $5000 with interest at 6% per annum 
has 5 years to run. It is necessary to realize on the mortgage. 
What sum should a person pay for it, if he wishes to make 7% 
on his money? 

Under what conditions would it be worth $5000? 



MISCELLANEOUS EXERCISES 301 

7. On November 1st, 1920, Brown invests $6000 in Victory 
Bonds, due November 1st, 1933, paying 5f% per annum 
payable half-yearly. If money is worth 6% per annum 
compounded half-yearly, what is the present worth of the 
bonds? (November 1st, 1921). 

8. A man deposits $200 a year with a loan company which 
pays 4% per annum compounded quarterly. What sum 
stands to his credit at the end of 5 years? 

9. A man dies leaving an annuity of $500 to his eldest son 
for 10" years and then to his second son for the following 10 
years. If money is worth 6% per annum, what is the present 
worth of each legacy? 

10. A municipality borrows $60,000 at 5% per annum. 
What amount must be collected each year so that the debt 
may be discharged in 10 equal annual payments, if money 
is worth 6% per annum? 

11. A man invests $500 in a business which pays 5% per 
annum. Each year he invests an amount 10% greater than 
the previous year. What amount stands to his credit at 
the end of 10 years, if he reinvests his dividends in the 
business? 

12. A man takes a 20-year endowment policy of $1000 on 
which the annual premium is $48-50. If he dies just after 
the twelfth payment, how much more will his heirs receive 
than if he had invested the money at 5% per annum? If he had 
lived, how much less will he receive than if he had invested 
the money as above? 

13. A man takes a straight life policy for $5000 on which 
the annual premium is $136. If he dies just before making 
the 25th payment, compare the financial returns with having 
invested the premium each year at 6%. 

MISCELLANEOUS EXERCISES. 

1. A man ordered 11^ tons of coal. The first load contained 
9500 lb., the second 7000 lb. How many tons remain to 
be delivered? 

2. A tank containing 400 gallons has two pipes opening 
from it. One pipe can empty it in 2 hours, the other in 2\ 
hours. If both^ pipes be opened for 15 minutes, how many 
gallons are left in the tank? 



302 MATHEMATICS FOR TECHNICAL SCHOOLS 

3. The recent summer vacation extended for 72 days. A 
boy spent ^ of it in the country, f of it camping, and the 
remainder in the city. How many days did he spend in the 
city? 

4. A bricklayer received 90c. an hour for an 8-hour day. 
In the last year he worked 221 days, what was his total income? 

5. A gang of men working on the roadway place 24| cu. 
yd. of concrete in 1 hour. How many cu. yd. do they place 
in 3 days of 8^ hours each? 

6. An alloy contains ^f copper, \ tin, and the balance zinc. 
How many lb. of each are there in an alloy of 336 lb.? 

7. If beef is worth 21f c. per lb., what is the value of beef 
weighing 532 lb.? 

8. One pipe can empty a tank in 3| hr., and another pipe 
can empty it in 2f hr. In what time can they both empty 
it if running together? 

9. A drill with a feed of -01" per revolution is making 
80 R.P.M. How long will it take to drill 25 holes through a 
\" plate if 15 seconds be lost in setting for each hole? 

10. A contractor is to finish a piece of roadway in 20 days. 
Twelve men work for 8 days and do \ of the work. How many 
men must be employed for the balance of the time to finish 
according to contract? 

11. A certain pump delivers 1-43 gallons per stroke. If a 
gallon of water weighs 10 lb., what weight of water will be 
delivered in 218 strokes? 

12. A drill with a feed of 100 is making 50 R.P.M. If \ 
of the time is used in setting, how many holes can be drilled 
in a \" plate in 2 hours? 

13. A gallon of water contains 277-274 cu. in. What is 
the percentage error in taking 6j gallons as equivalent to 1 
cu. ft.? 

14. A reamer 9" long is 1-375" in diameter at one end and 
1*125" at the other end. What is the taper per foot? 

15. How long will it take to excavate to a depth of 4' for 
a building having a frontage of 74' 0", and a depth of 243' 0", 
using a steam shovel if the bucket holds two-thirds of a cu. 
yd. and is filled 3 times every 2 minutes? 

16. The American gallon contains 231 cu. in. What per 
cent, is it of the English gallon? 



MISCELLANEOUS EXERCISES 303 

17. In building a roadway containing 1000 cu. yd., the 
total cost was as follows: — 4000 bags of cement at SI. 20; 
1000 cu. yd. of stone at $2.00; 450 cu. yd. of sand at $1.80; 
lumber $225; labor $3240, tools, etc., $200. Find the average 
cost per cu. yd. 

18. A pressure of 48-5 lb. per sq. ft. is how many ounces 
per sq. in.? 

19. A man walks 2\ miles. If his average step is 2' 7|", 
how many steps does he take? 

20. A room is to be floored with £" tongued and grooved 
flooring taken out of 6" material, allowing §* in width for 
the dressing and tongue. The room is 12' 6"X15' 6", and 
has a square bay window 10' 3" wide X 3' 0" deep. How 
many square feet are in the room and what number ot feet 
run of flooring would be required? 

21. Reduce 13 quarts to the decimal of a bushel. 

22. If sound travels 1120 ft. per sec, how long will it take 
to hear the report of a gun fired at a distance of 8| miles? 

23. An electric-light bill was 16f% higher this month than 
last. It was $3.42 last month, what is it this month? 

24. In a certain machine f of the power supplied is lost in 
friction, etc. What is the percentage efficiency of the machine? 

25. In a room 15' 3"X22' 6" it is required to lay a clear 
quarter cut white oak floor f'Xl^" face measure. Allowing 
30% for loss in dressing and working tongue, how much would 
the flooring cost at $275 per thousand sq. ft.? 

26. In a vernier caliper the reading shows 1", 2 tenths, 1 
small division, while the 8th division of the vernier is in line 
with a beam division. In reading the instrument I neglect to 
add the vernier reading. What is the percentage error? 

27. A steel hook when immersed in water displaced 1| 
quarts. If the specific gravity of steel is 7-8, find the weight 
of the hook. 

28. When casting iron pipe, an allowance of f " per foot is 
allowed for shrinkage. What percent, is this? 

29. If the specific gravity of ice is -921 and of salt water 
1-024, what part of an iceberg is below the surface of the 
water when floating? 

30. An alloy contains 22-5% of nickel and 1-2% of carbon. 
How many pounds of each are there in 500 pounds of the 
alloy? 



304 MATHEMATICS FOR TECHNICAL SCHOOLS 

31. Find the cost, at $60 per M, of building a walk 60' 
long by 6' wide; plank to be 2" thick and laid crosswise on 
3 pieces, 4"X4". 

32. A mixture for casting contains 5 parts copper, 4 parts 
lead, and 3 parts tin. What is the percentage composition? 

33. Find the cost of shingling a roof 16'X20' with shingles 
laid 4|" to the weather, if the cost of material and labour is 
$12 per square of shingles. 

34. A road bed rises 2' 3" in 300'. What percent, grade 
is this? 

35. Find the length of the . longest straight line that can 
be drawn in a room 20' long 16' wide and 12' high. 

36. A barn is 40' 0" wide and 60' 0" long, with a roof £ 
pitch. The rafter heel and projections at ends are each 2'. 
Find the cost of covering with 1 " sheeting at $60 per M, 
allowing 5% for waste. 

37. Find the weight of a hexagonal bar of iron |" to a side 
and 6' long. (1 cu. in. = -26 lb.). 

38. Find the size of tap drill for a &*, 12 pitch, sharp "V" 
thread nut. 

39. An equilateral triangle has an area of 32-54 sq. in. 
What is the length of a side? 

40. In a steel plate 4'X2' 3" and \" thick, 10 round holes 
are bored each \\" in diameter. Find the weight of the plate 
after boring. 

41. The diameter of the safety valve in a boiler is 3^". 
If the pressure of the steam is 150 lb. per sq. in., find the 
total pressure on the valve. 

42. A well is to be sunk 4' in outside diameter and 24' deep. 
If the carts used for carting away the earth hold 1^ cu. yd., 
and excavated earth increases in bulk 15%, how ninny cart 
loads will there be? A concrete slab 8" thick is placed on 
the bottom, and the sides are bricked, using 4 bricks per super- 
ficial foot. How many cu. yd. of concrete would be used 
and how many bricks? 

43. The driving wheels of a locomotive are 4' in diameter, 
and the speed of the locomotive is 40 miles per hour. How 
many revolutions must the drivers make per minute? 

44. A tank 7' in diameter is bound by 4 wrought-iron hoops 
2" wide and i§" thick. Find their weight. 



MISCELLANEOUS EXERCISES - 305 

45. The maximum speed for an emery wheel is a mile a 
minute. Find the maximum number of revolutions per sec. 
for an emery wheel 8" in diameter. 

46." A semi-circular platform with a diameter of 18' has a 
table 2' wide around its semi-circumference. Find the area 
of the table and the available space for seating accommodation. 

47. The length of the shadow of a 2' rod is 1' 6" and at the 
same time the shadow of a tree is found to be 30'. Find the 
height of the tree. 

48. What is the offset of the tail stock for turning a taper 
18" long on a bar 30" long if the diameters at the ends of the 
taper are 3$ " and 2\ " ? 

49. Find the size of the largest square timber which can be 
cut from a log 18" in diameter. 

50. Find the diameter of a tap drill for a Whitworth nut 
for a screw of outside diameter If", double threaded and 14 
pitch. 

51. Three circles each of radius 6" are enclosed in an equi- 
lateral triangle; find the side of the ^triangle. 

52. In riding a certain bicycle one revolution of the pedals 
gives two revolutions of the wheels. If the wheels are 26" 
in diameter, how many revolutions of the pedals per min. 
will give a speed of 12 miles per hour ? 

53. The diameter of a cylindrical winch barrel for a crane 
is 10". If a rope with a diameter of f " be used, how many 
coils of rope and what length of barrel would be necessary to 
raise a load 30' ? 

54. If the speed of a point on the circumference of a fly- 
wheel must not exceed 5000 ft. per min., find the maximum 
diameter for the wheel in order to make 120 R.P.M. 

55. A steel bar of square cross-section is to be equal in area 
to a rod 6" in diameter. Find a side of the square. 

56. A triangular steel plate is to have its sides 13", 14", 
and 15", and to weigh 2-98 lb.; find its thickness. 

57. An elliptical steel plate has a major axis of 12" and a 
minor axis of 10". It is f " thick; find its weight. 

58. A cylinder 14" in diameter fits in a cubical box. Cal- 
culate the percentage void. 

59. The commutator of a dynamo is 24" in diameter and 
16" long. Find the radiating surface in sq. ft. (lateral sur- 
face). 



306 MATHEMATICS FOR TECHNICAL SCHOOLS 

60. Find the diameter of a circular plate equal in area 
to an elliptical plate major and minor axis 18" and 12" 
respectively. 

61. The shaft of a square-headed bolt is 1" in diameter, 
the head being f " thick and l\" to the side. Determine the 
length of the shaft to have twice the weight of the head. 
(1 cu. in. = -26 lb.). 

62. Find to the nearest sixteenth of an inch the length of 
a |" steel rod that is turned per min., if the cutting speed is 
40 ft. per min. and the feed 24. 

63. The length of the core of a dynamo is 20" and it must 
have a radiating surface of 500 sq. in. Find its minimum 
diameter. 

64. An elliptical funnel has a major axis of 18' and a minor 
axis of 12'. Find the discharge of smoke in cu. ft. per min., 
if at a rate of 12 ft. per sec. 

65. The pulley on the armature shaft of a dynamo is 3\" 
in diameter. This is belted to a driving shaft which makes 
400 R.P.M. The speed of the dynamo must be 1800 R.P.M. 
What sized pulley must be placed on the driving shaft ? 

66. Find the weight of a coil of copper wire 400' long, 
if the area of the cross-section of the wire is 40,000 circular 
mils. (1 circular mil = area of a circle one mil, or -001", in 
diameter). 

67. Two hundred and forty-five sq. ft. of zinc are required 
in lining the sides and bottom of a cubical vessel. How 
many cu. ft. of water will it hold? 

68. Four concrete abutments are to be built for a bridge. 
Each abutment is to be 8' 0" high, 3' 0" thick, 18' 0" long at 
the bottom and 12' 0" at the top. If one cu. yd. of concrete 
requires 25 cu. ft. of stone, 12 cu. ft. of sand, and 4 bags of 
cement (1 bag=l cu. ft.), find the quantity of each necessary 
for the job. 

69. Find the total cost of building a rubble stone wall for 
a basement 30' 0" by 26' 0" by 8' 0" high, the wall being 18" 
thick, if the stone costs $60 a toise and the labour including 
sand and lime, is 30c. per cu. ft. (Exact length of wall is 
taken with no allowance for openings). 

70. It is required to build a brick wall on the front and one 
side of a lot 35' 0" in frontage and 128' 0" in depth. The 
wall is to be 7' 0" high and single brick 9" in thickness There 
are two gates one 12' 0"X7' 0" and the other 3' 0"X7' 0". 



MISCELLANEOUS EXERCISES 307 

Piers for strengthening the wall add 12% to its cubic contents. 
Allowing 15 bricks per cu. ft., at $18 per thousand, and $12 
per thousand for laying, find the total cost. 

71. A vessel to hold 10 quarts is to have an elliptical cross- 
section with major and minor axis 12" and 8" respectively. 
Find the height and the number of sq. ft. of tin in the vessel. 

72. A garage 10' 0" wide and 16' 0" long is to have a gable 
roof, | pitch. The following material is supplied: 

Rafters 2"X4", 2' on centre, 18" heel— $65 per M. 
Sheeting, 1" thick, projecting 6" on ends — $63 per M. 
Shingles laid A\ " to the weather at $9 • 50 per square. 
Find the cost of the above material. 

73. In drilling in mild steel a 1£" twist drill makes 40 
R.P.M., with a feed of 72. How many cu. in. will be cut 
away in 5 minutes? 

74. A chimney is to be built in a residence with the following 
dimensions: height from basement floor to first floor level is 
9' 0"; from first to second floor level 10' 6"; from second to 
third floor 9' 6"; and from third floor to top of stack 15' 0". 
The size of the stack from basement to second floor level is 
T 6" by 2' 7"; from second floor to third floor 4' 10" by 2' 7"; 
from third floor up 4' 1" by 2' 7". From basement to top of 
stack a 1' 1" by 1' 1" furnace flue is run. On first floor a 
fire-place opening 3' 0" by 2' 6" by 1' 10" deep is to be built. 
From top of fire-place opening a 9" by 1' 1" flue is carried up 
to top of stack. On the second floor a fire-place opening 
1' 10" by 2' 6" by 1' 1" deep is to be built having a 9" by 9" 
flue carried from top of opening to top of stack. Allowing 
15 bricks to the cu. ft., how many bricks would be necessary 
to build the stack ? 

75. Find the number of minutes required for turning a 
shaft 5" in diameter and 6' long, the cutting speed being 
40 ft. per min. and the feed 100. 

76. Two wheels 8" and 6" in diameter are running on parallel 
shafts 4' apart. Find the length of an open belt connecting 
the two wheels. (1) Using the formula S±(R + r)+2d. (2) 
Using the' exact method. Hence find the percentage error 
in the formula. 

77. The circumference of the base of a church spire in the 
form of a cone is 42' and the height is 80'. Find the cost of 
covering with sheet-iron at 30c. a sq. ft. 



308 MATHEMATICS FOR TECHNICAL SCHOOLS 

78. A chimney shaft 70' 0" high is to be erected having a 
flue averaging 3' 0"X3' 0" from bottom to top. The shaft 
is square and for the first 14' 0" the walls are 1' 10" thick; 
the next 14' 0" is 1' 6" thick; the next 20' 0" is 1' 1" thick 
and the remaining portion 9" thick. How many bricks 
would be required allowing 15 bricks to the cu. ft.? 

79. How many sq. ft. of sheet-iron will it take to roof a 
hemispherical dome 30' in diameter? 

80. A building 24' 0" wide and 36' 0" long is to have a hip 
roof, I pitch, with an 18" overhang, measured horizontally 
(formed by extending the rafters). Find the cost of the 
following material: 

Hip Rafters, 2 // X6' / -$55 per M. 

Rafters, 2"X6"(2' on centre), -§55 per M. 

Square sheeting, §" thick (10% added for cutting),— $53 

per M sq. ft. 
Slate, gauge 7\" , — $30 per square. 

81. A hollow copper sphere used as a float weighs 1 lb., 
and is 6" in diameter. How heavy a weight will it support 
in the water? 

82. Grain dumped in a pile makes an angle of 30° with the 
horizontal. How many bushels will there be if the pile forms 
a regular cone 10' in diameter? 

83. A tank 10' long and 2' in diameter is in the form of a 
cylinder with hemispherical ends. How many gallons will 
it hold? 

84. A steel pin 6" long and 1" in diameter at the large end 
has a B. & S. taper. Find its weight. 

85. A chimney shaft 80' 0" high is to be erected, having a 
flue averaging 3' 0" in diameter from. bottom to top. The 
shaft is circular and for the first 15' 0" the wall is 2' 0" thick, 
the next 15' 0" is 1' 6" thick, the next 25' 0" is 1' 0" thick, and 
the remainder is 9" thick. How many bricks will be required 
allowing 15 bricks per cu. ft.? 

86. An oil tank in the form of a cylinder 15' long and 3' 
in diameter is lying on its side. It is filled to a depth of 30". 
How many gallons of oil does it contain and what is the sur- 
face of the tank not in contact with the oil? 

87. A ring of outer diameter 16" is made of round cast- 
iron is" in diameter. Find its total surface and weight. 



MISCELLANEOUS EXERCISES 309 

88. A water pail has a base 12" in diameter and a top 16" 
in diameter. The height of the pail is 18". Find the capacity 
>n gallons and the sq. ft. of material used in construction. 

89. A spkere 8" in diameter is penetrated axially by a 
cylindrical hole 4" in diameter. Find the volume of the 
remaining solid. 

90. A tank is in the form of a cylinder with segments of 
spheres for ends. The total length is 8', the cylindrical part 
7', and the diameter 2'. Find the capacity in gallons. 

91. A building 24' 0" wide and 40' 0" long is to have a hip 
roof, ^ pitch, with a 2' overhang, measured horizontally (formed 
by extending the rafters). Find the cost of the following 
material : 

Hip rafters, 2" X6" -$60 per M. 

Rafters, 2" X6"(2' on centre), -$50 per M. 

Square sheeting, 1" thick (8% for cutting), — $45 per M. 

Shingles, 4j" to the weather, at $9.50 per square of shingles. 

92. In a room 16' 4" by 20' 8" it is required to lay a quarter 
cut clear white oak floor 3"X1§", face measure. Allowing 
30* { for loss in dressing and working tongue, find the cost at 
$250 per M square feet. 

93. A life-buoy elliptical in cross-section has major and 
minor diameters of 5" and 3" respectively. If the mean 
diameter be 30", find the volume in cu. in. 

94. A conical tent 9*' high is to be of such a size that a man 
6' high can stand erect anywhere within 3' of the centre pole. 
How many yd. of canvas 27" wide does it contain? 

95. Find the weight of a sheet of metal weighing 650 oz. 
per sq. ft., if the equidistant half ordinates at 1-5' intervals 
are 0, 1-75, 2-25, 3, 4-25, 6-35, 6 ft. 

96. Two wheels 9" and 8" in diameter are running on parallel 
shafts 5' apart. Find the length of a crossed belt connecting 
the two wheels. (1) Using the formula 3§(# + r)+2d. (2) 
Using the exact method. Hence, find the percentage error 
in the formula. 

97. A room 12' 0" wide and 17' 0" long is to be floored with 
No. 1 quality birch, |" X21", to cost 30c. a sq. ft. ; 33|% being 
added for dressing and working the tongue. It is also to be 
paneled to a height of 4' 0" at a cost of 80c. a sq. ft. There are 
2 doors each 2' 8" wide and 4 windows each 2' 6" wide, the sills 
being 2' 0" from the floor. Find the total cost. 



310 



MATHEMATICS FOR TECHNICAL SCHOOLS 



98. A lot 50' 0" X160' 0" is to be enclosed by a picket fence. 
The pickets are 4' 0" long, 3" wide and 1" thick, and are placed 
3" apart. The posts are placed 8' 0" apart, scantling 2" X 
4" being used at top and bottom for railing, and a base-board 
10" wide. If the posts cost 30c each, the lumber $52 per M, 
and the pickets $10 per hundred, find the total cost of material. 

99. With a feed of \" per revolution, how fast is it neces- 
sary to run a bar, to turn 40" long in 10 minutes? 




Fig. 131 

100. The above figure represents a cross-section of an egg- 
shaped sewer. OE is the right bisector of AB and equal in 
length to AB. The semi-circular top has a radius 0.4 = 12". 
The sides are arcs with radii CB and DA each equal 36". 
The small end is an arc with radius FE = Q>". Find the area 
of the cross-section. 



MISCELLANEOUS EXERCISES. 

1. The area of a circle is irr 2 ; express the diameter in terms 
of the area. 

2. Express the area of a rectangle in terms of its perimeter 
when the length is twice the width. 



MISCELLANEOUS EXERCISES 311 

3. Three circles are to touch one another and have their 
centres 3", 4" and 5" apart. Find the diameters of the 
circles. 

W ird 2 

4. The stress in a tie bar is /=-r where A = -j~. Find 

d when/ = 7000 and JF = 2100. 

5. The difference of the acute angles of a right-angled 
triangle is 15°. Find the angles. 

6. The weight of a body varies inversely as the square of 
its distance from the centre of the earth. If a man weighs 
190 lb. at the earth's surface, what would he weigh on the 
top of a mountain 4£ miles high (radius of earth = 4000 miles). 

4 

7. The volume of a sphere is -^t 3 . Find the radius when 

o 

the volume is 616 cu. in. 

A 

8. The stress in a beam is given by/ = ,— . Find A when/ = 

cd 2 
6000, «=p c = 3, d = 5. 

9. The time of vibration of a pendulum varies as the square 
root of its length, for a given latitude. A pendulum 39 • 1" 
long vibrates once per second. What is the length of a pendu- 
lum that vibrates three times per second? 

10. An exterior angle at the base of an isosceles triangle 
is 108°; find all the angles. 

11. The velocity of flow of water under a head h is V — 
ay/2gh. Find h when V = 36, g = 32, a = • 5. 

12. In an isosceles right-angled triangle the perp. from the 
vertex on the hypotenuse is 8", find all the sides. 

13. The twisting moment on a solid circular shaft is given 
by T = ^-. Find d when T= 150000, / = 6000. 

14. A chord 8" long is 6" from the centre of a circle. Find 
the radius of the circle. 

15. The moment of inertia of a body about an axis is given 

7 2 

by I = — . Find k when I = • 5, w = 38, a = 32. 
9 

16. The radius of a circle is 16". How far from the centre 
is a chord 8" in length? 



312 MATHEMATICS FOR TECHNICAL SCHOOLS 

17. The law of a machine is given by P = x+yw. Find x 
when P = 6-48, y= -2, w = 62. 

18. The length and breadth of a rectangular floor differ 
by 6'; the area is 72 sq. ft., find the perimeter. 

19. An equilateral triangle has sides 20" in length. Find 
the radius of the inscribed and circumscribed circles. 

20. The coefficient of self-induction of a coil of wire is 

. T 4ttAti 2 
given by L= /1Q> . 

Find n, when A = irr 2 , r = 2-5, Z=-015, Z = 40. 

21. The difference between the lengths of the parallel sides 
of a trapezium is 4; the area is 100, and the sum of the parallel 
sides 20. Find the dimensions. 

22. The lifting power of an electro-magnet is given by 

8x ' 
P being the pull in dynes. Find P, when B = 14000, A = 30. 

23. I have to walk a distance of 144 miles, and I find that 
if I increase my speed by 1^ miles per hour I can walk the dis- 
tance in 14 hours less than if I walk my usual rate. Find my 
usual rate. 

24. In measuring a rectangle the length is measured \\% 
too small and the width 2% too large. Find the percentage 
error in the area. 

25. The perpendicular from the vertex of the right angle 
of a right-angled triangle to the hypotenuse is 2§". The 
hypotenuse is 5". Find the other sides. 

26. The length of a cylinder is twice its diameter. Find 
the diameter so that the cylinder may contain three times as 
many cu. ft. as a sphere 6" in diameter. 

27. A ton of lead is rolled into a sheet \ " thick. Find the 
area of the sheet if a cu. ft. of lead weighs 712 lb. 

28. A rectangular piece of tin has an area of 195 sq. in., 
and its perimeter is 56". Find its dimensions. 

29. Around the outside of a square garden a path 3' wide 
is made. If the path contains 516 sq. ft., find a side of the 
garden. 



MISCELLANEOUS EXERCISES 313 

30. An open box is made from a square piece of tin by 
cutting out a 2" square from each corner and turning up the 
sides. How large is the original square if the box contains 
1152 cu. in.? 

31. A man whose eye is 5' 6" above the ground, sights over 
the top of a 12' pole and just sees the top of a tower. If he is 
7' from the pole and 63' from the tower, find the height of 
the tower. 

32. A derrick for hoisting coal has its arm 24' long. It 
swings over an opening 20' from the base of the arm. How 
far is the top of the arm above the opening? 

33. Show that the area of a triangle is y/s (s — a) (s — b) (s — c) , 
where a, b, and c are the sides and s half their sum. 

34. An open box is made from a rectangular piece of tin 
twice as long as it is wide, by cutting out a 2" square from each 
corner and turning up the sides. If the total surface is 56 
sq. in., find the dimensions of the original piece. 

35. At a school entertainment the price of the tickets 
for the second performance was reduced 20%, which resulted 
in an increase in receipts of 10%. What was the percentage 
increase in the number of tickets sold ? 

36. Two branches of an iron water pipe are respectively 
\\" and 2\" in diameter. Find the diameter of a pipe that 
will just carry away the water from both branches. 

37. If a man spent \ of his salary for board, \ of the remainder 
for other expenses, and saved annually $400, what was his 
salary? 

38. The resistance offered by the air to the passage of a 
bullet through it varies jointly as the square of its diameter 
and the square of its velocity. If the resistance to a bullet 
whose diameter is -25" and whose velocity is 1600' per second 
is 48-5 oz., what will be the resistance to a bullet whose 
diameter is -4" and whose velocity is 1550' per second? 

39. The sides of the base of a triangular prism are as 3:4:5, 
and its volume is 270 cu. in. If the altitude is 5", find the 
sides of the base. 

40. The ends of a frustum of a cone are respectively 8" 
and 2" in diameter. If the lateral surface is equal to the area 
of a circle whose radius is 5", find the height of the frustum. 



314 MATHEMATICS FOR TECHNICAL SCHOOLS 

41. The safe distributed load in wood beams is given by 

c\by.d 2 
S.L. = j , where c is a const., b the breadth in in., 

d the depth in in., L the length in ft. 
Solve for c, b, d and L. 
If c = 100, 6 = 10", d = 6", Z = 10' find S.L. 

42. The horse-power required to drive air through a pipe 

Q 3 L 
is given by H.P. = J~ „ „ , where Q is the volume in cu. ft. per 

sec., L the length in ft., d the diameter in in. 
Solve for Q, L and d. 
If Q = 10 cu. ft., L = W, d = 10", find H.P. 



TABLES 



315 



Decimal Equivalents of Parts of an Inch. 



67' 

1 
32 .... . 

3 

"ST 

1 
TTT 

5 

■gT' 

3 
-3~2' • ' • 

7 

"BT 

1 

¥ 

9 

"BT 

5 
-3~2' 

11 

3 

TB 

13 
64 

7 

T2- • • • 

15 
"BT 

T 

17 

"BT 

19 
64 

A 



•01563 

•03125 

•04688 

•0625 

•07813 

•09375 

• 10938 
•125 

• 14063 

• 15625 
•17188 
•1875 
•20313 
•21875 
•23438 
•25 

•26563 
•28125 
•29688 
•3125 



21 

64 



A3. 

3 2 



A5. 
32 



TS' 



19 
32 



21 

32' 



23 
64 



25 

6 1 



2J7 
64 



29 
^4 



31 
64 



33 
64 



35 
64 



37 
64 



39 
64 



4JL 

64 



43 

64 



•32813 

•34375 

•35938 

•375 

•39063 

•40625 

•42188 

•4375 

•45313 

•46875 

•48438 

•5 

•51563 

•53125 

•54688 

•5625 

•57813 

•59375 

•60938 

•625 

•64063 

•65625 

•67188 

•6875 



45. 

6 4 • 

2 3 

32 

47 
64 • 

3 

T 

49 
64 • 

3 2 

51 

64 ' 

1 3 

16 

_51 
6~4 • 

2 7 

32 

55 
64 • 

T 

8 

5.1. 
64 • 

29 

32 

_5j>. 
64 • 

1 5 

16 

6 1 • 

3 1 

32 

63 
64 • 

1 1 



•70313 
•71875 

• 73438 
•75 

•76563 
•78125 
•79688 
•8125 
•82813 
•84375 
•85938 
•875 
•89063 
•90625 
•92188 
•937.5 
•95313 
•96875 
•98438 

• 00000 



316 



MATHEMATICS FOR TECHNICAL SCHOOLS 



The following Tables are from Kent's Engineers' Pocket 
Book: 

Weight and Specific Gravity of Stone, Brick, Cement, Etc. 





Specific 
Gravity 


Weight in lbs. 
per cubic foot 


Asphaltum 


1-39 
1-6 

1-79 
2-0 
216 
2-24 to 2-4 
218 
1-6 
1-79 


87 


Brick, Soft 

" Common 


100 
112 


" Hard 


125 


" Pressed 


135 


" Fire 


140 to 150 


" Sand-Lime 


136 


Brickwork in Mortar 

" in Cement ....... 

Portland Cement (loose) .... 


100 

112 

92 


(in barrel) . 
Clay 




115 


1-92 to 2-4 
1-92 to 2-48 
1 • 15 to 1 • 28 
1-44 to 1-76 
2-56 to 2-72 
1-6 to 1-92 
•8 to -96 
2-3 to 2-9 
2-56 to 2-88 
2-24 to 2-56 
2-24 to 2-68 
1-44 to 1-6 

115 
1-50 to 1-81 

2-64 
1-44 to 1-76 
2-24 to 2-4 
2-72 to 2-88 
2-16 to 3-4 
1-76 to 1-92 


120 to 150 


Concrete 


120 to 155 


Earth, loose 


72 to 80 


" rammed 


90 to 110 


Granite 


160 to 170 


Gravel 


100 to 120 


Lime, Quick, in bulk 

Limestone 


50 to 60 
140 to 185 


Marble 


160 to 180 


Masonry, dry rubble 

M dressed 


140 to 160 
140 to 180 


Mortar 


90 to 100 


Pitch 


72 


Plaster of Paris 


93 to 113 


Quartz 


165 


Sand 


90 to 100 


Sandstone 


140 to 150 


Slate 


170 to 180 


Stone, various 


135 to 200 


Tile 


110 to 120 







TABLES 



317 



Weight and Specific Gravity of Wood. 





Specific 

Gravity 
Mean Value 


Weight in lbs. 
per cubic foot 


Ash 

Bamboo 

Beech 


1 


72 
35 
73 
65 
62 
66 
56 
53 
23 
61 
59 
38 
81 
68 
77 
74 
45 
61 
48 
45 
82 
58 
54 


45 
22 
46 


Birch 


41 


Cedar 


39 


Cherry 


41 


Chestnut 

Cypress 


35 
33 


Ebony 


76 


Elm 


38 


Fir 


37 


Hemlock 

Mahogany 

Maple 


24 
51 
42 


Oak, White 

Oak, Red 

Pine, White 

Pine, Yellow 

Poplar 


48 
46 
28 
38 
30 


Spruce 


28 


Teak 


51 


Walnut 


36 


Willow 


34 











318 MATHEMATICS FOR TECHNICAL SCHOOLS 

Weight and Specific Gravity of Metals. 



Aluminum 

Brass : — Cu. + Zn 
'80 20^ 
70 30 



60 
,50 



40 
50J 



Bronze 

Copper 

Iron, Cast .... 
Iron, Wrought 

Lead 

Magnesium. . . 

Mercury 

Nickel 

Platinum 

Silver 

Steel 

Tin 

Tungsten 
Zinc 



Specific 

Gravity 

Mean Value 



2-67 

8-6 

8-40 

8-36 

8-2 

8-853 

8-853 

7-218 

7-7 

11-38 
1-75 

13-6 
8-8 

21-5 

10-505 
7-854 
7-35 

17-3 
7-00 



Weight in lbs. 
per cubic foot 



166-5 



Weight in lbs. 
per cubic inch 



0963 



536-3 


•3103 


523-8 


•3031 


521-3 


■3017 


511-4 


•2959 


552 


•3195 


552 


•3195 


450 


•2604 


480 


•2779 


709-7 


•4106 


109 


•0641 


848 


•4908 


548-7 


•3175 


1347 


•7758 


655 • 1 


•3791 


489-6 


•2834 


458-3 


•2652 


1078-7 


•6243 


436-5 


•2526 



LOGARITHMS 
Logarithms. 



319 



























Mean Differences 







1 


2 


3 


4 


5 


6 


7 


8 


9 
















1 2 


3 


4 


5 6 


7 8 9 


10 


ooooi i 


00432 


00860 


01284 


01703 












42 85 127 


170 212 254 


297 339 381 














02119 


02531 


02938 


03342 


03743 


40 81 


121 


162 202 242 


283 323 364 


II 


04139 


04532 


04922 


05308 


05690 












37 77 116 


154 193 232 


270 309 348 














06070 


06446 


06819 


07188 


07555 


37 74 111 


148 185 222 


259 296 333 


12 


07918 


08279 


08636 


08991 


09342 












36 71 


106 


142 177 213 


248 284 319 














09691 


10037 


10380 


10721 


11059 


34 68 102 


136 170 204 


238 272 307 


13 


11394 


11727 


12057 


12385 


12710 












33 66 


98 


131 


164 197 


229 262 295 














13033 


13354 


13672 


13988 


14301 


32 63 


95 


120 


158 190 


221 253 284 


14 


14613 


14922 


15229 


15534 


15836 












30 61 


91 


122 


152 183 


213 244 274 














16137 


16435 


16732 


17026 


17319 


29 59 


88 


118 147 177 


206 236 265 


15 


17609 


17898 


18184 


18469 


18752 












28 57 


86 


114 142 171 


199 228 256 














19033 


19312 


19590 


19866 


20140 


28 55 


83 


110 


138 165 


193 221 248 


16 


20412 


20683 


20951 


21219 


21484 












27 53 


80 


107 134 160 


187 214 240 














21748 


22011 


22272 


22531 


22789 


26 52 


78 


104 


130 156 


182 208 233 


17 


23045 


23300 


23553 


23805 


24055 












26 50 


70 


101 


126 151 


176 201 227 














24304 


24551 


24797 


25042 


25285 


25 49 


73 


OS 


122 147 


171 196 220 


18 


25527 


25768 


26007 


26245 


26482 












24 48 


71 


95 


119 143 


167 190 214 














26717 


26951 


27184 


27416 


27646 


23 46 


00 


03 


116 139 


102 185 208 


19 


27875 


28103 


28330 


28556 


28780 












23 45 


0s 


00 


113 135 


158 180 203 














29003 


29226 


29447 


29667 


29885 


22 44 


00 


88 


110 132 


154 176 198 


20 


30103 


30320 


30535 


30750 


30963 


31175 


31387 


31597 


31806 


32015 


21 43 


64 


85 


106 127 


148 170 190 


21 


32222 


32428 


32634 


32838 


33041 


33244 


33445 


33646 


33846 


34044 


20 40 


00 


80 


100 120 


140 160 180 


22 


34242 


34439 


34635 


34830 


35025 


35218 


35411 


35603 


35793 


35984 


20 39 


58 


77 


97 116 


135 154 174 


23 


36173 


36361 


36549 


36736 


36922 


37107 


37291 


37475 


37658 


37840 


19 37 


50 


74 


93 111 


130 148 167 


24 


38021 


38202 


38382 


38561 


38739 


38917 


39094 


39270 


39445 


39620 


18 35 


53 


71 


89 106 


124 142 159 


25 


39794 


39967 


40140 


40312 


40483 


40654 


40824 


40993 


41162 


41330 


17 34 


51 


68 


85 102 


119 136 153 


26 


41497 


41664 


41830 


41996 


42.160 


42325 


42488 


42651 


42813 


42975 


16 33 


40 


66 


82 98 


115 131 148 


27 


43136 


43297 


43457 


43616 


43775 


43933 


44091 


44248 


44404 


44560 


16 32 


47 


63 


79 95 


111 126 142 


28 


44716 


44871 


45025 


45179 


45332 


454S4 


45637 


45788 


45939 


46090 


15 30 


40 


(il 


76 91 


107 122 137 


29 


46240 


46389 


46538 


46687 


46835 


46982 


47129 


47270 


47422 


47567 


15 29 


44 


50 


74 88 


103 118 132 


30 


47712 


47857 


48001 


48144 


48287 


48430 


48572 


48714 


48855 


48996 


14 29 


43 


57 


72 86 


100 114 129 


31 


49136 


49276 


49415 


49554 


40003 


49831 


49969 


50106 


50243 


50379 


14 28 


41 


55 


69 83 


97 110 124 


32 


50.51.5 


50050 


50786 


50920 


51054 


51188 


51322 


51455 


51587 


51720 


13 27 


40 


54 


67 80 


94 107 121 


33 


51851 


51983 


52114 


52244 


52375 


52504 


52634 


52763 


52892 


53020 


13 26 


30 


52 


65 78 


91 104 117 


34 


53148 


53275 


53403 


53529 


53656 


53782 


53908 


54033 


54158 


54283 


13 25 


38 


50 


63 76 


88 101 113 


35 


54407 


54531 


54654 


54777 


54900 


55023 


55145 


55207 


55388 


55509 


12 24 


37 


4!) 


61 73 


85 98 110 


36 


55630 


55751 


55S71 


55991 


56110 


5022!) 


5634s 


56467 


56585 


56703 


12 24 


36 


48 


60 71 


83 95 107 


37 


56S20 


56937 


57054 


57171 


57287 


57403 


57519 


57634 


57749 


57864 


12 23 


35 


40 


58 70 


81 93 104 


38 


57978 


58092 


58206 


58320 


58433 


58546 


58659 


58771 


58883 


58995 


11 23 


34 


45 


57 68 


79 90 102 


39 


59106 


59218 


59329 


59439 


59550 


59660 


59770 


59879 


59988 


60097 


11 22 


33 


44 


55 66 


77 88 99 


40 


60206 


60314 


60423 


60531 


60638 


60745 


60853 


00050 


61066 


61172 


11 22 


33 


44 


55 66 


77 88 99 


41 


61278 


61384 


61490 


61595 


61700 


61805 


61909 


62014 


62118 


62221 


10 21 


31 


42 


53 63 


74 84 95 


42 


62325 


62428 


62531 


62634 


62737 


02 S3 


62941 


63043 


63144 


63246 


10 20 


31 


41 


51 61 


71 82 92 


43 


03347 


63448 


63548 


63649 


63749 


63846 


63949 


64048 


64147 


64246 


10 20 


30 


40 


50 60 


70 80 90 


44 


64345 


64444 


64542 


64640 


64738 


64836 


64933 


65031 


65128 


65225 


10 20 


20 


39 


49 59 


68 78 88 


45 


65321 


65418 


65514 


65510 


65706 


65801 


65896 


65992 


66087 


66181 


10 19 


20 


88 


48 57 


67 76 86 


46 


66276 


66370 


66464 


0055s 


66652 


66745 


66839 


66932 


67025 


67117 


9 19 


28 


37 


47 56 


65 74 84 


47 


67210 


07302 


67394 


67486 


67578 


67669 


67761 


67852 


67943 


68034 


9 18 


27 


30 


46 55 


64 73 82 


48 


68124 


6821S 


68305 


6S395 


68485 


68574 


68664 


68753 


68842 


68931 


9 18 


27 


36 


45 53 


63 72 81 


49 


69020 


69108 


69197 


6928S 


011373 


69461 


6954s 


69636 


69723 


69810 


9 18 


20 


35 


44 53 


62 70 79 



320 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Logarithms. 

























Mean Differences 







1 


2 


3 


4 


5 


6 


7 


8 


9 








1 2 3 


4 5 6 


7 8 9 


50 


69897 


BOOM 


70070 


70157 


70243 


70829 


70418 


70501 


70688 


70672 


9 17 26 


34 43 52 


60 69 77 


51 


707.57 


70842 


70927 


71012 


71096 


71181 


71265 


71349 


71433 


71517 


8 17 25 


34 42 50 


59 67 76 


52 


7 1'IDi i 


71684 


71767 


71850 


71933 


72016 


72099 


72181 


72263 


72346 


8 17 25 


33 42 50 


58 66 75 


53 


72428 


72509 


72591 


72673 


72754 


72835 


72916 


72997 


73078 


73159 


8 16 24 


32 41 49 


57 65 73 


54 


73239 


73320 


73400 


73480 


73560 


73640 


73719 


73799 


73878 


73957 


8 16 24 


32 40 48 


56 64 72 


55 


74036 


74115 


74194 


74273 


74351 


74429 


74507 


74586 


74663 


74741 


8 16 23 


31 39 47 


55 63 70 


56 


74819 


74896 


74974 


75051 


75128 


75205 


75282 


75358 


75435 


75511 


8 15 23 


31 39 46 


54 62 69 


57 


76687 


75664 


75740 




75891 


75967 


76042 


76118 


76193 


76268 


8 15 23 


30 38 45 


53 60 68 


58 


76343 


76418 


76492 


76567 


76641 


76716 


76790 


76864 


76938 


77012 


7 15 22 


30 37 44 


52 59 67 


59 


77085 


77159 


77232 


77305 


77379 


77452 


77525 


77597 


77670 


77743 


7 15 22 


29 37 44 


51 58 66 


60 


77815 


77887 


77960 


78032 


78104 


78176 


78247 


78319 


78390 


78462 


7 14 22 


29 36 43 


50 58 65 


61 


78533 


78604 


78675 


78746 


78817 


78888 


78958 


79029 


79099 


79169 


7 14 21 


28 36 43 


50 57 64 


62 


79239 


79309 


79379 


79449 


79518 


79588 


79657 


79727 


79796 


79865 


7 14 21 


28 35 41 


48 55 62 


63 


79934 


80003 


80072 


80140 


80209 


80277 


80346 


80414 


80482 


S0550 


7 14 20 


27 34 41 


48 54 61 


64 


80618 


80686 


80754 


80821 


80889 


80956 


81023 


81090 


81158 


81224 


7 13 20 


27 34 40 


47 54 60 


65 


81291 


81358 


81425 


81491 


81558 


81624 


81690 


81757 


81823 


81889 


7 13 20 


26 33 40 


46 53 59 


66 


81954 


82020 


82086 


82151 


82217 


82282 


82347 


82413 


82478 


S2543 


7 13 20 


26 33 39 


46 52 59 


67 


82607 


82672 


82737 


82802 


82866 


82930 


82995 


83059 


83123 


83187 


6 13 19 


26 32 38 


45 51 58 


68 


83251 


83315 


83378 


83442 


83506 


83569 


83632 


83696 


83759 


83822 


6 13 19 


25 32 38 


44 50 57 


69 


83885 


83948 


84011 


84073 


84136 


84198 


84261 


84323 


84386 


84448 


6 12 19 


25 31 37 


43 50 56 


70 


84510 


84572 


84634 


84696 


84757 


84819 


84880 


84942 


85003 


85065 


6 12 19 


25 31 37 


43 50 56 


71 


85126 


85187 


85248 


85309 


85370 


85431 


85491 


85552 


85612 


85673 


6 12 IS 


24 31 37 


43 49 55 


72 


85733 


85794 


85854 


85914 


85974 


86034 


86094 


86153 


86213 


86273 


6 12 18 


24 30 36 


42 48 54 


73 


86332 


86392 


86451 


86510 


86570 


86629 


86688 


86747 


86S06 


86S64 


6 12 18 


24 30 35 


41 47 53 


74 


86923 


86982 


87040 


87099 


87157 


87216 


87274 


87332 


87390 


87448 


6 12 17 


23 29 35 


41 46 52 


75 


87506 


87564 


87622 


87679 


87737 


87795 


87852 


87910 


87967 


88024 


6 12 17 


23 29 35 


41 46 52 


76 


88081 


88138 


88195 


88252 


88309 


88366 


88423 


88480 


88536 


88593 


6 11 17 


23 29 34 


40 46 51 


77 


88649 


88705 


88762 


88818 


88874 


88030 


88086 


89042 


89098 


89154 


6 11 17 


22 28 34 


39 45 50 


78 


89209 


89265 


89321 


89376 


89432 


89487 


89542 


80507 


89653 


89708 


6 11 17 


22 28 33 


39 44 50 


79 


89763 


89818 


89873 


89927 


89982 


90037 


90091 


90146 


90200 


90255 


6 11 17 


22 28 33 


39 44 50 


80 


90309 


90363 


90417 


90472 


90526 


90580 


90634 


90687 


90741 


90795 


5 11 16 


22 27 32 


38 43 49 


81 


90848 


90902 


90956 


91009 


91062 


91116 


91169 


91222 


91275 


91328 


5 11 16 


21 27 32 


37 42 48 


82 


91381 


91434 


91487 


91540 


91593 


91645 


91698 


91751 


91803 


91855 


5 11 16 


21 27 32 


37 42 48 


83 


91908 


91960 


92012 


92064 


92117 


02169 


92221 


92273 


92324 


92376 


5 10 16 


21 26 31 


36 42 47 


84 


92428 


92480 


92531 


92583 


92634 


92686 


92737 


92788 


93840 


92891 


5 10 15 


20 26 31 


36 41 46 


85 


92942 


92993 


93044 


93095 


93146 


93197 


93247 


93298 


93349 


93399 


6 10 15 


20 26 31 


36 41 46 


86 


93450 


93500 


93551 


93601 


93651 


93702 


93752 


93802 


93852 


93902 


5 10 15 


20 25 30 


35 40 45 


87 


93952 


94002 


94052 


94101 


94151 


94201 


94250 


94300 


'M.m 


94399 


5 10 15 


20 25 30 


35 40 45 


88 


94448 


94498 


94547 


94596 


94645 


94694 


94743 


94792 


04841 


94890 


5 10 15 


20 25 29 


34 39 44 


89 


94939 


94988 


95036 


95085 


95134 


95182 


95231 


95279 


95328 


95376 


5 10 15 




34 39 44 


90 


95424 


95472 


95521 


95569 


95617 


95665 


95713 


95761 


95809 


95856 


5 10 14 


19 24 29 


34 38 43 


91 


95904 


05062 


95999 


96047 




96142 


96190 


96237 


96284 


96332 


5 9 14 


19 24 28 


33 38 42 


92 


B637S 


96421 


96473 


06630 


96567 


96614 


96661 


96708 


96755 


96802 


5 9 14 


19 24 28 


33 38 42 


93 


9884S 


96V.I.- 


96942 


06988 


97035 


97081 


07128 


97174 


97220 




5 9 14 


18 23 28 


32 38 42 


94 


97313 


97351 


97405 


97451 


07497 


97543 


97589 


97635 


07681 


97727 


5 9 14 


18 23 28 


32 37 42 


95 


97772 


9781* 


97864 


9790S 


979.').". 


98000 


98046 


98091 




98182 


5 9 14 


18 23 27 


32 36 41 


96 


98227 


98275 


98318 




08408 




08408 


98543 


98588 




5 9 14 


18 23 27 


32 36 41 


97 


08677 


08725 


08767 


98811 


• 


98900 


08046 


08089 




99078 


4 9 13 


is 22 27 


31 36 40 


98 


0012; 


00167 


00211 


00251 


9 


09344 


0038J 


00432 




99520 


4 9 13 


18 22 2i 


31 35 40 


99 


09664 


0060; 




99693 


09739 


097 S3 


00831 


99870 




99957 


4 9 13 


17 22 21 


31 35 39 



ANTILOGARITHMS 



321 



Antilogarithms. 









2 


3 


4 


5 


6 


7 


8 


9 


Mean Differences 







1 
































1 


2 3 


4 5 6 


7 8 9 


.00 


10000 


10023 


10046 


10089 


10093 


10116 


10139 


10162 


10186 


10209 


2 


5 7 


9 12 14 


16 19 21 


.01 


10233 


10257 


10280 


10304 


10328 


10351 


10375 


10399 


10423 


10447 


_) 


5 7 


10 12 14 


17 19 21 


.02 


10471 


10495 


10520 


10544 


10568 


10593 


10617 


10641 


10666 


10691 


2 


5 7 


10 12 15 


17 20 22 


.03 


10715 


10740 


10765 


10789 


10814 


10839 


10864 


10889 


10914 


10940 


3 


5 8 


10 13 15 


18 20 23 


.04 


10965 


10990 


11015 


11041 


11066 


11092 


11117 


11143 


11169 


11194 


3 


5 8 


10 13 15 


18 20 23 


.05 


11220 


11246 


11272 


11298 


11324 


11350 


11376 


11402 


11429 


11455 


3 


5 8 


11 13 16 


18 21 24 


.06 


11482 


1150s 


11535 


11561 


11588 


11614 


11641 


11668 


11695 


11722 


3 


5 8 


11 13 16 


19 21 24 


.07 


11749 


11776 


11803 


11830 


11858 


11885 


11912 


11940 


11967 


11995 


3 


5 8 


11 14 16 


19 22 25 


.08 


12023 


12050 


12078 


12106 


12134 


12162 


12190 


]221s 


12246 


12274 


3 


6 8 


11 14 17 


20 22 25 


.09 


12303 


12331 


12359 


12388 


12417 


12445 


12474 


12503 


12531 


12560 


a 


6 9 


11 14 17 


20 23 26 


.10 


12589 


12618 


12647 


12677 


12706 


12735 


12764 


12794 


12823 


12853 


3 


6 9 


12 15 18 


21 24 26 


.11 


12882 


12912 


12942 


12972 


13002 


13032 


13062 


13092 


13122 


13152 


3 


6 9 


12 15 18 


21 24 27 


.12 


13183 


13213 


13243 


13274 


13305 


13335 


13366 


13397 


13428 


13459 


3 


6 9 


12 15 18 


21 25 28 


.13 


13490 


13521 


13552 


13683 


13614 


13646 


13677 


13709 


13740 


13772 


3 


6 9 


13 16 19 


22 25 28 


.14 


13804 


13836 


13868 


13900 


13932 


13964 


13996 


14028 


14060 


14093 


3 


6 10 


13 16 19 


22 26 29 


.15 


14125 


14158 


14191 


14223 


14256 


14289 


14322 


14355 


14388 


14421 


3 


7 10 


13 16 20 


23 26 30 


.16 


14454 


14488 


14521 


14555 


14588 


14622 


14655 


14689 


14723 


14757 


3 


7 10 


13 17 20 


24 27 30 


.17 


14791 


14825 


14859 


14894 


14928 


14962 


14997 


15031 


15066 


15101 


3 


7 10 


14 17 21 


24 28 31 


.18 


15136 


15171 


15205 


15241 


15276 


15311 


15346 


15382 


15417 


15453 


4 


7 11 


14 18 21 


25 28 32 


.19 


15488 


15524 


15560 


15596 


15631 


15668 


15704 


15740 


15776 


15812 


4 


7 11 


14 18 22 


25 29 32 


.20 


15849 


15885 


15922 


15959 


15996 


16032 


16069 


16106 


16144 


16181 


i 


7 11 


15 18 22 


26 30 33 


.21 


16218 


16255 


16293 


16331 


16368 


16406 


16444 


16482 


16520 


16558 


4 


8 11 


15 19 23 


26 30 34 


.22 


16596 


16634 


16672 


16711 


16749 


16788 


16827 


16866 


16904 


16943 


4 


8 12 


15 19 23 


27 31 35 


.23 


16982 


17022 


17061 


17100 


17140 


17179 


17219 


17258 


17298 


17338 


4 


8 12 


16 20 24 


28 32 36 


.24 


17378 


17418 


17458 


17498 




17579 


17620 


17660 


17701 


17742 


4 


8 12 


16 20 24 


28 32 36 


.25 


17783 


17824 


17865 


17906 


17947 


17989 


18030 


18072 


18113 


18155 


4 


8 12 


17 21 25 


29 33 37 


.26 


18197 


18239 


18281 


18323 


18365 


18408 


18450 


18493 


18535 


18578 


4 


8 13 


17 21 25 


30 34 38 


.27 


18621 


18664 


18707 


18750 


18793 


18836 


18880 


18923 


18967 


19011 


4 


9 13 


17 22 26 


30 35 39 


.28 


19055 


1909! 


19143 


19187 


19231 


19275 


19320 


19364 


19409 


194.54 


4 


9 13 


18 22 26 


31 35 40 


.29 


19498 


19543 


19588 


19634 


19679 


19724 


19770 


19815 


19861 


19907 


5 


9 14 


18 23 27 


32 36 41 


.30 


19953 


19999 


20045 


20091 


20137 


20184 


20230 


20277 


20324 


20370 


5 


9 14 


19 23 28 


32 37 42 


.31 


20417 


20464 


20512 


20559 


20606 


20654 


20701 


20749 


20797 


20845 


5 


10 14 


19 24 29 


33 38 43 


.32 


20893 


20941 


20989 


21038 


21086 


21135 


21184 


21232 


21281 


21330 


fi 


10 15 


19 24 29 


34 39 44 


.33 


21380 


2142S 


21478 


21528 


21577 


21627 


21677 


21727 


21777 


21827 


5 


10 15 


20 25 30 


35 40 45 


.34 


21878 


21928 


21979 


22029 


22080 


22131 


22182 


22233 


22284 


22336 


5 


10 15 


20 25 31 


36 41 46 


.35 


22387 


22439 


22491 


22542 


22594 


22646 


22699 


22751 


22803 


22856 


■> 


10 16 


21 26 31 


37 42 47 


.36 


22909 


22961 


23014 


23067 


23121 


23174 


23227 


23281 


23336 


23388 


5 


11 16 


21 27 32 


37 43 48 


.37 


23442 


23496 


23550 


23605 


23659 


23714 


23768 




23878 


23933 


5 


11 16 


22 27 33 


38 44 49 


.38 


23WS8 


24044 


24099 


24155 


24210 


24266 


24322 


24378 


24434 


24491 


6 


11 17 


22 28 34 


39 45 50 


.39 


24547 


24604 


24660 


24717 


24774 


24831 


24889 


24946 


25003 


25061 


6 11 17 


23 29 34 


40 46 51 


.40 


25119 


25177 


25236 


25293 


25351 


25410 


25468 


25527 


25586 


25645 


6 12 18 


23 29 35 


41 47 53 


.41 


25704 


25763 


25823 


25882 


25942 


26002 


26062 


26122 


26182 


26242 


6 


12 18 


24 30 36 


42 48 54 


.42 


26303 


26363 


26424 


264& 


26546 


26607 


26669 


26730 


26792 


26853 


6 


12 18 


24 31 37 


43 49 55 


.43 


26915 


26977 


27040 


27102 


27164 


27227 


27290 


27353 


27416 


27479 


6 13 19 


25 31 38 


44 50 56 


.44 


27542 


27606 


27669 


27733 


27797 


27861 


27925 


27990 


28054 


28119 


6 13 19 


26 32 39 


45 51 58 


.45 


28184 


28249 


28314 


28379 


28445 


28510 


28576 


28642 


28708 


28774 


7 13 20 


26 33 39 


46 52 59 


.46 


28840 


28907 


28973 


29040 


29107 


29174 


29242 


29309 


29376 


29444 


7 13 20 


27 34 40 


47 54 60 


.47 


29512 


29580 


29648 


29717 


29785 


29854 


29923 


29992 


30061 


30130 


7 14 21 


28 34 41 


48 55 62 


.48 


30200 


30259 


30339 


30409 


30479 


30549 


30620 


30691 


30761 


30832 


7 


14 21 


28 35 42 


49 56 63 


.49 


30903 


30974 


31046 


31117 


31189 


31261 


31333 


31405 


31477 


31550 


7 


14 22 


29 36 43 


50 58 65 



322 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Antilogarithms. 

























Meu 


Differences 









1 


2 


3 


4 


5 


6 


7 


8 


9 














1 2 3 


4 


5 


6 


7 


8 9 


.50 


31623 


31696 


31769 


31842 


31916 


31989 


32063 


32137 


32211 


32285 


7 15 22 


29 


37 


44 


52 


59 66 


.51 


32358 


32434 


3250! 


32584 


3265! 


32735 


3280! 


32SS5 


32001 


33037 


8 15 23 


30 


38 


45 


53 


60 68 


.52 


33113 


33189 


33266 


33343 


33420 


33497 


33574 


33051 


33729 


33806 


8 15 23 


31 


30 


4( 


54 


62 69 


.53 


33884 


3301)3 


34041 


34110 


34198 


34277 


34356 


34435 


34514 


34594 


8 16 24 


32 


40 


47 


55 


63 71 


.54 


34074 


34754 


34834 


34914 


34995 


35075 


35156 


35237 


35318 


35400 


8 16 24 


32 


40 


48 


56 


65 73 


.55 


35481 


35563 


35645 


35727 


35810 


35892 


35975 


36058 


36141 


36224 


8 16 25 


33 


41 


51 


58 


66 74 


.56 


36308 


36392 


36475 


30559 


36044 


36728 


36813 


30898 


36983 


37068 


8 17 25 


34 


42 


51 


50 


68 76 


.57 


37154 


37239 


37325 


37411 


37497 


37584 


37070 


37757 


37844 


37931 


9 17 26 


35 


43 


52 


61 


69 78 


.58 


38019 


38107 


38194 


38282 


38371 


38459 


38548 


38637 


38726 


38815 


9 18 27 


35 


44 


53 


62 


71 80 


.59 


38905 


38994 


39084 


39174 


39264 


39355 


39446 


39537 


39628 


39719 


9 18 27 


36 


45 


M 


63 


72 82 


.60 


39811 


39902 


39994 


40087 


40179 


40272 


40365 


40458 


40551 


40644 


9 19 28 


37 


46 


56 


65 


74 83 


.61 


40738 


40832 


40920 


41020 


41115 


41210 


41305 


41400 


41495 


41591 


9 19 28 


38 


47 


57 


66 


76 85 


.62 


41687 


41783 


41879 


41970 


42073 


42170 


42267 


42364 


42462 


42560 


10 19 29 


39 


40 


58 


68 


78 87 


.63 


42658 


42756 


42855 


42954 


43053 


43152 


43251 


43351 


43451 


43551 


10 20 30 


40 


BO 


60 


70 


80 89 


.64 


43652 


43752 


43853 


43954 


44055 


44157 


44259 


44361 


44463 


44566 


10 20 30 


41 


51 


61 


71 


81 91 


.65 


44668 


44771 


44875 


44978 


45082 


45186 


45290 


45394 


45499 


45604 


10 21 31 


42 


52 


02 


73 


83 94 


.66 


45709 


45814 


45920 


40020 


46132 


40238 


46345 


46452 


46559 


46666 


11 21 32 


43 


53 


64 


75 


85 . 96 


.67 


46774 


46881 


46989 


47008 


47200 


47315 


47424 


47534 


47643 


47753 


11 22 33 


44 


54 


65 


76 


87 98 


.68 


47863 


47973 


48084 


48195 


48300 


48417 


48529 


48641 


48753 


48865 


11 22 33 


45 


56 


67 


78 


89 100 


.69 


48978 


49091 


49204 


49317 


49431 


49545 


49059 


49774 


49888 


50003 


11 23 34 


46 


57 


68 


80 


91 103 


.70 


50119 


50234 


50350 


50466 


50582 


50699 


50816 


50933 


51050 


51168 


12 23 35 


47 


58 


70 


82 


93 105 


.71 


51286 


51404 


51523 


51642 


51701 


51880 


52000 


5211!! 


52240 


52360 


12 24 36 


48 


60 


72 


84 


96 108 


.72 


52481 


52602 


52723 


52845 


52000 


53088 


53211 


53333 


53456 


53580 


12 24 37 


40 


61 


73 


85 


98 110 


.73 


53703 


53S27 


53951 


54075 


54200 


54325 


54450 


54570 


54702 


54828 


13 25 38 


n 


63 


75 


88 


100 113 


.74 


54954 


55081 


55208 


55336 


55403 


55590 


55719 


55847 


55976 


56105 


13 26 38 


51 


64 


77 


90 102 115 


.75 


56234 


56364 


56494 


56624 


56754 


56885 


57016 


57148 


57280 


57412 


13 26 39 


52 


66 


70 


92 105 118 


.76 


57.544 


57677 


57810 


57943 


58070 


58210 


58345 


58470 


58614 


58749 


13 27 40 


54 


67 


80 


94 


107 121 


.77 


58884 


50020 


59150 


59203 


50420 


50560 


59704 


50841 


59979 


00117 


14 27 41 


:>:> 


69 


82 


96 110 123 


.78 


60256 


60395 


00534 


60674 


60S14 


00954 


61004 


01235 


61376 


61518 


14 28 42 


56 


70 


84 


98 


112 126 


.79 


61659 


61802 


01944 


62087 


62230 


62373 


62517 


62061 


62806 


62951 


14 29 43 


58 


72 


86 


101 


115 130 


.80 


63090 


63241 


63387 


63533 


63680 


63S26 


63973 


64121 


0420!) 


64417 


15 29 44 


50 


74 


ss 


103 118 132 


.81 


Ii4565 


64714 


04803 


65013 


65163 


65313 


05404 


65615 


05706 


65917 


15 30 45 


60 


75 


00 


105 


120 135 


.82 


66069 


66222 


06374 


66527 


00081 


60834 


00988 


67143 


67298 


67453 


15 31 46 


62 


77 


01' 


108 


123 130 


.83 


67608 


67764 


67020 


68077 


08234 


68301 


686«i) 


68707 


68865 


69024 


16 32 47 


03 


70 


95 


110 126 142 


.84 


69183 


09343 


69503 


69663 


69823 


69984 


70140 


70307 


70469 


70632 


10 39 48 


64 


81 


07 


113 


129 145 


.85 


70795 


70958 


71121 


71285 


71450 


71614 


71770 


71945 


72111 


72277 


17 33 50 


(,^ 


83 


89 


110 


132 149 


.86 


72444 


72611 


72778 


72046 


73114 


73282 


73451 


73621 




73961 


17 34 51 


68 


86 


101 


118 


135 152 


.87 


74131 


74302 


74473 


74645 


74S17 


74989 


75162 


75336 


75500 


75683 


17 35 52 


69 


87 


104 


121 


138 156 


88 


75858 


76033 


76208 


76384 


76560 


76736 


76013 


770!K) 


77268 


77446 




71 


80 


107 


125 


142 150 


.89 


77625 


77804 


77983 


78163 


78343 


78524 


78705 


78886 


79068 


70250 


18 36 54 


72 


01 


100 


127 


145 163 


.90 


79433 


79616 


79799 


79983 


80168 


80353 


80538 


S0724 


80910 


81096 


19 37 56 


74 


93 


111 


130 


148 167 


.91 


81283 


81470 


81658 


81846 


82035 


82224 


82414 


82604 


82794 


82985 


19 38 57 


76 


95 113 


132 


151 170 


.92 


Vi 171. 


S330S 


83560 


83753 


83046 


84140 




84528 


81723 


84918 


19 39 58 


78 


97 


116 


136 


155 175 


93 


S5114 


85310 


85507 


85704 


85901 


86099 


86298 


86497 


86696 


80896 


20 40 60 


70 


99 


11!) 


130 


158 178 


.94 


87096 


87297 


87498 


87700 


87902 


88105 


88308 


88512 


88716 


88920 


20 41 61 


81 


102 


122 


142 


162 183 


.95 


89125 


89331 


89586 


89743 


89950 


90157 


90305 


90573 


90782 


90991 


21 42 62 


83 


101 


125 


146 


166 187 


.96 


.11201 


91411 


91022 


91833 


92045 


02257 


92470 


92083 


02897 


93111 


21 42 64 


85 


106 


127 


149 


170 191 


.97 


13325 


03541 


93756 


03072 


04180 


01 106 


94624 


94842 


96060 


95280 


22 43 65 


87 


109 


130 


152 


174 105 


.98 


15499 


05710 


05040 


06161 




96605 


96828 


97051 


07275 


97499 22 44 67 


39 


11 


133 


155 


178 200 


.99 


'77:' 1 


07010 


98175 


98401 


0S628 


ivs.V, 




99312 


0054 1 


00770 23 40 68 


>1 


14 


137 


lik) 


182 205 



ANSWERS 



Page 

4.— -7. 11413. £.23437. 5.16438. 4-392-26. 5.365-09. 
6. 1907-80. 7. 79-07. 8. 1158-94. 9. 2220-65. 

10. 8380-882. 11. 35880-920. 12. 10964-8526. 
13. $1056-95. U- $2221-39. 15. $2253-15. 

5.— 1. 7535. £.2-36. 5.364-947. 4-139-20.5.34408-79. 

6. 3-207. 7. 103-30. 8. 2-4116. 9. 11-232. 10. 

217-209. 
7.—1. 1557. 2. 1309. 3. 6152. 4- 952. 5. 55368. 

6. 2960. 7. 34500. 5. 764-5. 9. 7864. 10. 11-800. 
11.1788. 12.17880. 15.7864. 14. -0469. 15.4-69. 
16. 4290-5. 17. 18408. 15. 78. 19. 76-14. 00. 1180. 

21. 236896000. 22. 2538. 25. 1018400. 2k. 6885. 
5.— 1. 3276. 2. 6344. 5. 4796. 4- 31433. 5. 67392. 6. 3213. 

7. 1910520. 8. 2809566. 9. 5049668. 10. 699678. 
11.494-34. 12.26430-588. 15.1299-276. 14- 69-765. 
15. -021112. 16. -038934. 17.28-58392. 18. -7916832. 
19. 50092-640436. 20.. 1-3255716. 21. 11-109280. 

22. 1-47504. 23. 78-028125. 24- 2112-280092. 
25. 95-797296. 26. 3-38928. -27. 2971-6736. 28. 
19-25625. 29. -0000625. 30. 1-55697696. 

10.— 1. 32987. 2. 180912. 5. 1-764. 4- 11^- 5. -00158. 
6. 1-76. 7. -01325. 8. 1-22. 9. -025. 10. 85-1. 

11. 17-02. 12. 31196i 13. 46-42. 

75.— 1. 65-02. 2. 64. 5. 679-65. 4- 2119-92. 5. 134-29. 

6.278-10. 7.3-04. 5.7-97. 9.14-19. 10.4071-17. 

11. 297-99. 12. 1-033. 
lk.—l. 18. 2. 20. 5. 44. 4. 70. 5. 6891. 6. 585. 7. 183. 

15.— 1. 28. 2. 3f. 5. 9. 4- 240000. 5. 230^. 6. ^A- ?• t¥t- 

8. 2ff 9. 8. 10. 48. 11. 76-8. 1& 115-2. 13. 2-5. 
14- 22$. 1. 1133 lb. 

323 



324 MATHEMATICS FOR TECHNICAL SCHOOLS 

Page 

16.— £.22-9175. 5.252 ft. 4-3199. 5.3773,3507. 6.465481b. 

267721b. 7.2361b. 5.144. 9. 14 rem. 2 ft. 10. 177$ miles. 

11.60. 12.251 miles. 13. $10287-16. 14.2400. 
17.— 15. 8. 16. 72. 

Of) 1 6 O 20 <? 4 5 /. 21 K 1 5 /? 4 9 7 102 O 6 8 

Q 32 If) 72 

*• ^5"^"- • /t7 - rr&"- 

£>/) 7 1 O X .?i /, ^ T2 /J 3 7 7 242 O 107 

Q 9 rn 5 

t7 ' 100- - 117, 16- 

21.— 1. 3£. £. 4. 5. 7$. 4- 6. 5. 8f. 6. 17f 7. 29$. 5. 3ft. 

9. 11 T 2 T . 10. 27^. 11. 2500. 12. 9-&. 15. 53. 14. 9. 

15. V°- 16- ts- I 7 - ff- ^- H¥-- I 9 - H 1 - 20. *$*. 

£1. S-Jg 5 - 8 . flg. 1 ^2o 21 - ^- -w 3 - H> ^g^. 
22.— 1. H- *• If- 5. A- 4- Iff- 5. $ff. 6. f$. 7. 1^. 5. Iff. 

9. $f. 10. 11|. 11. 3ff- i*. 3^fr. 15. 19ft- *4- 8Hf. 
£5.-1. 33f lb. 0. |f. 5. 2^". 4, 2$f ". 5. 24 T 5 F . 6. 19$*. 
£4.-7. 3f*. 5. 16ft', 11*. 9. 2A', 2f , Iff. ^- If; 
26.— 1. 2f. £. 2f. 5. if. 4- A- 5. ft. 6. A. 7. / T . 5.^. 

9. If, 10. f$. 11. M$*. 12. 1 ¥ V 15. 500. 14. 4f. 

15. 62. 16. 25. 17. 3. 15. 1. 19. 5^ T . £0. 39ff. JM. 49. 
22. 4f. £5. 1. 2Jf. 27ff 

27.—1.-&. 2. ft. 3. ^. 4. A- 5.^. 6. if. 7. 3f. 8.376$. 
9.6. 10.^. 11. 7ff. 12.$$. 15. 12fff. ** riV. 15. 1^. 

16. 16$f 

©7 7 4 3 £> 1 <? 94 /, 3567 K 251 fi 1 7 14 1 

*'• l - T0T7- *• ITS'- J - TZS- 4- 50^0- °- 507- °- T27- '■ IToo- 
O 617 Q 2 1/1 31 

°- "S~ooTT- a - TT5- - LU - 200000- 

28.— 1. -25. 0. -5. 5. -375. 4- -6875. 5. -6. 6. -992. 7. -9G875. 

5. -9375. 9. -96. 10. -9921875. 11. -74. 1#. -508. 
29.— 1. -5. 0. -083. 5. -i42857. £ -13. 5. -05. 6. • is. 

7. -307692. 

50.-1. f. 2. A- *.**• -WiV 5-tVs- S-tVt- 7.3$f- * ff$- 

Ql /rt 9 23 7/0589 1® 78 7 7<3 11 7/ 7 1 

tf. T . iU. Z^q. 11. ^ TT xo- <**• 166 5 0- IO ' 3 00 0- J ^- T¥T5"- 

M.— *. 4, 2, 22i 60. 5. 50%, 25%, 12f %. 4- 50%, ? 

33*%. 5. 416|%, 4$$%. 6. 15, 35, 70, 85. 7. 80. 5. 70. 
5£.— 9. $8-07£. 10. 30 gallons. 



ANSWERS 325 

Page 

33.— 1. 10, 5, 10. 2. $78-30. 3. 11^ hr. 4- $167-47. 

5. first by 4|c. 6. 21-2 lb. 7. $72-25. 5. 9ff. 9. 4. 

10. $840-00. 11. $1692-90. 1£. $3125-00. 13. $25631-41. 

r /, 4 2 1 

54.— i5. ff- in. 16. i 17. 8f. IS. 10 lb. 19. $27-03f. 

20. 73. 21. $8-44. &g. 54-94 lb. copper, 27-06 lb. zinc. 

23. 9 T Vr days. #4- $360-00. £5.63-295. £6.309-462. 

27. 232| oz. £3. 6 T 6 g in. 
35.— 29. 55-97". 50.83-27. 31.4-5225". 3£. 96. 33. 83f. 

34. 78. 35. -25, $4000. 36. 1062-5, 187-5. 37. 14 in. 

38. 822-64 cu. in. 39. 72. 40. 8000 lb. 41. 11§ in. 
36.— 4£. If in. 43. $5775-00. 44- $4100-00, $6150-00, 

$6150-00. 45. 107^, 35f£%. 46. 65f. 47. 10383-36. 
38.— 1. 1760. 0. 5280. 3. -027. 4. -003125. 5. 681". 

6. 16339^'. 7. .000125. 3. -00126. 9. 415 li. 10. l^&ch. 

11. 2874-96'. 12. ^ ch. 13. 52-37 miles. 

39.-5. 14', 4-4406". 6. 9525 cm., -09525 dm. 7. -06096 Km. 
40— 8. 981-456 cm. 9. 1-34 Km. 10. 134- 112 cm. 
41.— 4. 4840 sq. yd. 5. 1321 sq. ft. 6. -033 ac. 7. 6930 sq. in. 

3. -51 sq. yd. 9. 2} ac. 10. -803 sq. ft. 1J. ff sq. ft. 

1£. 198. 13.96. 14.572V 15. 2-16 ac. 
42.— 16. 24012. 17. $373-33. 13. S6-53. 19. 72. 20. 54ff. 
43.-3. 225 cu. ft. 4- 143432 cu. in. 5. 6790-363 cc. 6. Iff 

7. 3^oV -37. 3. 21^y, 104. 9. $10-67. 10. $143-44. 
11. 129. 

44-—12. 442-45. 13. 19' 6". 14- $112-12, $128-34. 15. \". 

16. 36300. 
46.— 1. 58-35 cu. ft. 2. 1-83. 
47.— 3. 1-03. 4- 90864-018 tons. 5. -79. 1.196. 2. 484 days. 

3. 45563 min. 4- 1496. 
49.-3. 343 1, 343000 g. 4- 3875 cu. in., 1096-625 lb". 5. 2268 

Kgm. 6.52310. 7.6271-98. 3.12145-14. 9. 3600| lb., 

°2T- 

51.— 1.45. £.199. 3.123. 4-327. 5.37. 6.1-732. 7.3-4908. 
3.11-9668. 9. -7370. 10. -2507. 11.220yd. 12. 11-67". 
13. 30-59". 14. The square pipe. 



326 MATHEMATICS FOR TECHNICAL SCHOOLS 

Page 

52.— 15. 15-87". 1. 14-14". 0.48-82'. 5.51-38'. 4.127-27'. 

5. 469-04'. 

53—6. 121-96'. 7. 20-85'. 

5k.— 1. 6-39. 0. 5-93. 3. (a) $159-05. (6.) $146-33. 

55— U- 228-97. 572-42, 1144-48. 5. 2809f 

56.— 0. 21409. 5.30491. 4. $220-20, $313-63. 5.66248. 

6. $2647-17. 

59.-5. 6, 72. 6. 13$, 160. 7. $143-60. 8. $139-78. .9. $44-69. 
65.— 2. $5-46, $11-76, $24-49. 5. $8-00, $14-93, $34-99. 
66.— J,. $25-00, $33-86, $52-33. 5. $95-63, $114-55, $496-40. 

6. $47-12. 

67.-7. $64-83. 1. (a) $57-02, (6) $88-59, (c) $197-50. 
0. (a) $91-04, (6) $92-16, (c) $173-40. 3. $180-21. 

68.-4. $375-28. 5. $800-42. 

69.— 1. $6-97, $10-39, $8-25. 

70.— 2. $4-97, $2-20, $10-20. 3. $10-11, $8-99, $8-50. 

89.— 1. 70', 14'. 0; 320. 3. 620'. £. 160. 5. 5760 sq. ft. 

90.— 6. 1501". 7.120'. 8.12. 9. $6000-00. 10. 4800 cu. ft., 
1200 cu. ft, 11. 15'. 12. 18", 6". 13. 48c, 32c, U. 65c, 
26c 15. 2 years. i6. 3 days. 17. 9 T V min. 2£. $f day. 

91.— 19. 25 miles, 30 miles. 00. 16f. 0i. $200000. 22. 68 T 2 T . 
05. 2100 gals. 2A. $68750-00. 25. $18000-00. 26. 4£' 
from fulcrum. 27. 10 cm. 

93.— 11. 19 T V 20. -tV **• 4 !§- *•*• If- 

9^.— 1. -2a. 0. 2a-3o-3c. 5. -8x + 23?/+3z. 

4. 7sy-yz+10zx. 5. 3z 2 + 3x-6. 6. 7a. 7. 6x 3 -2x 2 
-*+2. 8. 4# 3 + 2 / 2 + 3. 

102.—l.x+±. 2. a + l. 3. a-2. 4. *-"-7. 5. 3x-4. 6. 3-2a. 

7. 2+z. 5. s+y. 9. 5-3a. 10. 2x 2 + 7. ii. z+b. 
12.x 2 +xy+y 2 . 13.a 2 -ab+b 2 . U. x*+Zz % y+%xy*+y t . 
15. a 2 -\-b 2 -\-c 2 — ab — bc — ca. 

104.— 2. \, 88, 37-5. 3. -0315, 47-7, 238-1. & 26775, 480, £. 

5. 15, 3. 6. 8-jSj-, 73i 2i 55, 110. 



ANSWERS 327 

105.-7. 26-832, 24-4, -000327. 5. 1-04, -024. 9. 2318-4, 14. 

10. -448, 254-375, 20-56, 115-5. 11. 65-625, -375, 

40533|, -75, 78-75, 10-83. 
106.— 12. 4-42, 22-38, 186-5. 13. 1584, 34-72, 52 09, H. 300, 

20, 25. 15. 4-76, 33-98, 34-375, 4033-3. 
107.— 16. 34f, 291§, 4|, 171-81. 
113.— 3. 6, 1-81, 139-4. 
114.— 5. H ft. 6. 72. 7. $258-00. 8. 192 sq. in. 9. 10 lb. 

10. 57-12. 11. $157-28. 12. $404-60. 

115.— 13. $10-80. 14. $9-73. 15. -7936 acres. 16. 5H sq. ft. 

17. 1-37218, V3 half the base. 19. 23-382 sq. in. 

20. 13-93 sq. in. 21. 15. **. 60. 05. 22-4. 24- 111". 
116.— 25. 10-825 sq. in., 62-352 sq. in., 73-177 sq. in. 

06. 8-75 sq. in. 27. 41 sq. in. 28. 77 sq. ch. 09.3-75 
sq.ft. 30. 312 acres. 31. 142-5 sq. ft. 50. -6605. 

117.— 33. 7-2703. 54- 3-4208. 55. 1777. 

118.— 36. 158-1. 57. 14-9', 9-4', 828^ sq, ft. 55. 8912 sq. ft. 

105.— 2. 131 -95". 5.4-71'. 4-4618,52-48. 5.59-9". 6.240. 

7. 3769-92. 5. 1884-96. 9. 15-28". 10. 18041-8 miles. 

11. 14". 10. 140, 360. 13. 300 R.P.M. 14- 13". 
124.— 18. 314-16, 64403 lb. 19. 9-63 sq. in. 

125.— 20. The 4" pipe. 01. 4, 9. 00. 5-383". 05. 12-36". 

24. 15-6". 06. 14-92 sq. in. 07. 87-965 sq. in. 

05. 6387-27 sq. yd. 29. 34-56 sq. in. 30. 7-61 sq. in. 

51. $9414-91. 50. 20-11 sq. ft. 
126.— 33. 18-7". 54. 525 sq. in. 35. 41-33 sq. ft. 36. 67-2 

sq. in. 
132.— 3. 549-78', 23562 sq. ft. 4- 44-11 lb. 5. -22". 6. 28". 

7. -649 sq. in. 8. 13-25. 
133.— 9. 123-11 sq. ft. 10. -373 lb. 11. 10-454'. 10. -866 

. sq.ft. 16. 598-113 sq. in. 17. 202- 1 sq. ft, 
157.-1.4:3. 0.5:2. 5.3:5. 
155.-4.168,112. 5.335:6.6.20:49. 7. 196 T 4 T . 5.175. 9. $144. 

10. $56-25. 11. llf 10. $53-34. 15. 83-3'. 14. 51f. 

15. 54'. 16. 124, 93, 31. 17. 60, 24, 12. 



328 MATHEMATICS FOR TECHNICAL SCHOOLS 

Page 

139.-18.4^'. 19. 28£". 20: ^^ 21. 15'. 22. 11,80'. 28. 1056'. 

U3.—1. 1200, 800. 2. \ , - 5 T °. 3. 66§, 133J, J. 5|, lOf. 
5. T \, 5, P= T VW+5. 6.6000,20. 7. -046, 5. 5.21-6, 
•0044. 9. 1957, 1521. 10. 345, 245. 

1U.—11. 100, 200. 12. 4-&, 5J. 13. 106-15, 72-21. U. 10-42, 
•0088. 15. $21000000, $1890000. 1. 1-35, 3-26, 32. 
2. 9-4, 15. 

U5.—3. 102, 210, -2-13, 14. *. 1190-4, 20, 10. 5. -548, 106f, 
63, 21504, -658. 6. 30062- 1, -377, -65. 

146.— 7. 13-4, 12-2, 7, 16. 8. 6283200, -31416, -079. 9. -58, 
1125, -7. 10. 17-2, 161-6, 24-8. 11. 111-2, 217, 
•06,6-66. 

U7.—12. 171", 44-5". 13. 245-25", 31-5". 1-4- 6-66, 1069-2. 
15. 18-6", 34-9, 4400. 16. 11", 58-18, 3300. 

W.— 17. 42 -215, 35 -9, 6 -9. 18. 1-0472, 14-3, 86450. 19.667-6, 
18390, 149-21. 

W.— £0. 3, 1181-9. 
179.— 1. 19-64. 0. 68-75. 

180.— 3. 63-63. 4- 392-85. 5. 22-92. 6. 9-55". 7. 23-86. 
8. 11-45". 9. 2357 + . 10. 76-36. 

181.— 1. 2304. £. 9-60 min. 3. 16 min. 4- 11-31 min. 
5. #f. 6. 5 min. 7. 0028". S. 5 min. 9. 79-2. 

10. 33-94. 

187.— 1. |". 2. 1". 3. -8625". 4- 12". 5. 5". 6, 7^". 
7. No. Morse. 8. B. & S. 9. -604". 20. No. Mq 

11. \". 12. Jarno. 1-4- if". 15. £". 16. ^" . 17. T %". 
188.-18. (a) |f", (6) T y, (c) \", (<£) f ". 1.9. If. 50. -4". 

£1. 3^", tV'. 22. 2-1". 23. 1-68". ££. 2° .V.)'. 
2° 52' 24", 2° 23', 2° 51' 50". 

191.— 2. 12. 5. 10. 4- H- 5- H- «• H- 7- *• *■ ; '-- 
9.1". 10.1-615". 

193— 1. -1625". 5. -4541". 3. A- -4- 4". 5. A- 6. 4. 






7. 3. S. U. 9. *. 10. * 



ANSWERS 329 

Page 

195.— 1. 1-187". 2. 2-167". 5. |. 4- 3". 6. If 7. |. 

5. 4. 9. 6. 

i97._i. -05336", -01144". 2. fr, -62193". 5. -08004", 
•83992. 4- -0915", \, -0196". 5. 4£, 1 -7154", -02948". 

6. -04", -295", -00858". 

202.— 1. 24, 64. 0. 32, 48. 3. 45. 4- 16. 5. 24, 72. 6. 36. 

7. i 5. 24. 9. 72. 10. 42, 98. 

205.— 1. 4\. 0. 24 stud, 92 lead, 36 inside c, 72 outside c. 
206.— 3. 48 stud, 28 lead, 36 inside c, 72 outside c. 4- 24 stud, 

96 lead, 72 inside c, 36 outside c. 5. 42 stud, 98 lead. 

6. 24 stud, 96 lead. 7. 64 stud, 40 lead. 5. 112. 9. Equal 

gears. 10. 36 inside, 72 outside. 
210.— 1. 8". 2. -2618". 3. -1122". 4- -1541". 5. -0982". 

6. -1348". 7.5-166". 5.5-5". 9.70. 10.46. 11.10. 
12. 7". 

212.-1.4:19. 2.40. 5.4-09". 4.42. 5.30-4. 6.509". 

7. f" per min. 5. 2" per min. 
213.— 9. 5. 10. i". 

225.-2. 15". 4- 15". 5. 21-333". 6. 21-5". 7. 49° 57'. 
226.-8. 22° 48'. 9. 1". 10. 14-945". 1. J». 2. £. 3. 10. 
4. -1443". 5. 9. 6. -3466", *. 7. 9f min. 5. 7l£. 

9. \", 2° 23'. 10. |". 11. 5-487". 12. T V 13. 28. 
227.— 14- 8. 15.72. 16.36. 17. llj. 18. 5. 19. 36 inside, 

72 outside. 20. 6-5". 21. 11". 08. 3-82". 25. 6f. 
2k- 24 worm, 32 second stud, 64 first stud, 72 screw. 
25. 33° 8'. 
239.— 1. 3-0099. 2. -4420. 5. -08836. 4- -89827. 5.7-5229. 
6. 21-015. 7. 3-3332. 5. -58798. 9. -0055873. 

10. -0007237. 11. 5800-9. 12. 4-4419. 13. 1-2057. 
14.1-878. 15. -2999. 16. -25507. 17.2-37. 15.1-80. 
19. 1-68. 

240.— 20. 279-04 sq. yd. 21. 188-86 sq. in. 22. 43-301 
acres. 25.13-221. 24.475-4'. 25.50-77". 26.16495 
sq.in. 27.30-73. 25.13-365. 29.40101 + . 50.592-21. 



330 MATHEMATICS FOR TECHNICAL SCHOOLS 

Page 

242.-2. $27-20. 5. 68-35. 4- 472. 5. 10-57 sq. ft., 

2-29 cu. ft. 6. 5-41. 
2U-—3. 128-18 lb. 4. 11846+. 5. $26-18. 
245.-6. 28 • 49 lb. 2. 502 • 65 sq. in. 
246.-3. 667 • 98 lb. 4- 2127 • 89 lb. 5. 522 • 78 lb. 6. • 280 lb. 

248.-2. 48 sq. in., 11-46 sq. in. 3. 50-03 lb. 4- 1615 1b. 

5. (a) $71-97, (b) 22200 + . 

249.-6. 70-86. 

250.— 2. 31-196 lb. 5. 131-4 sq. in., 24-11 lb. 4- 7856-4, 
$49-74. 5. 3-59 cu. in. 

254.— ». 91 sq. in. 3. 496-61 lb. 4- 48-01 lb. 5. 30-73 cu. in. 

6. 2-36 cu. in. 

256.-3. 67-02. 4- 41-34 lb. 5. 1-55 lb. 6. 2-29 lb. 

257.— 1. 907 sq. ft. fc 611-3 sq. ft. 

258.-3. 74-22 sq. ft., 41-86 cu. ft. 4- 352-19 sq. in., 490-09 
cu. in. 5. 837-76 cu. in. 

259.— 1. 30 1b. 2. 2988 tons. 3. 3325-64 lb. 4- 7854 1b. 
5. 165-93. 6. 10102+ lb. 1. 1-41". 2. 821-6 lb. 

260.— 3. 5 12". 4- -153/ 5.16-53'. 6.2 02 1b. 7. 70-68 lb. 
8. 25-01'. 9. 3291-48 sq. ft. 10. 4-84. 11. 22-53. 
12. 51-30 sq. in. 13. 20 min. 14. 13675+ lb. 
15. 603 19 sq. in. 16. $17-00. 17. 163-17. 

261.— 18. 25-79. 19. 2787 cu. in. 20. $255-00. 21. 26-51 lb. 

22. 8774-30 lb. 23. -243. 24- 2-65". £5. 22-15 lb. 

26. 5-01". 07. 11581 + . 28. 648000. 09. 1942 1 sq. 

ft., 10603+ cu. ft. 30. $73-67. 
262.— 31. 12-57 cu. in. 50.3-32". 55. 19215+ lb. 5^. 79-79. 

55. 376 -99 sq.m. 36. 4 -69 lb. 57.14-50". 55.13-76. 

39. 12-84". 40. 9-59". 
264.— 11. (a-b)(x-y). 12. (x + z)(x-y). 13. (3 + a)(x-y). 

14- 0r 2 -?/)(.T-2). 15. (ax- b) (bx- a). 16. (a 4 + l)(a + l). 

17. (a 2 -6)(l+c). 15. (a +3) (0a 2 - c). 19. (a: + m 2 )(x+m). 

00. (a + b){x + y-z). 



ANSWERS 331 

Page 

267.-5. (a+b + c)(a-b-c). 6. {x 2 +y 2 ){x+y){x-y). 

7. (x*+y*)(x 2 + y 2 )(x + y)(x-y). 8. (a + b + c)(a-b-c). 

9. (x + y + a + b)(x + y-a-b). 10. (x + y)(x+y)(x-y) 
(x-y). 

268.— 11. (x+y-z)(x-y + z). 12. (l-a-6)(l + a + 6). 

13. (a 8 + l)(a 4 + l)(a 2 + l)(a + l)(a-l). 

Ik. (x-y + a + b)(x-y-a-b). 17. 5(a-6+2c) 
\a-b-2c). 18. (4-a + 6)(4 + a-6). 

19. (2x 2 +xy+3y 2 )(2x 2 -xy+3y 2 ). 

20. (x 2 -3x + 9){x 2 +3x + 9). 

21. (x 2 + 2xy + 2y 2 )(x 2 -2xy+2y 2 ). 

22. (x 2 -x-l)(x 2 +x-l)(x i +Sx 2 + l). 

23. (2x + y)(2x-y)(x-3y)(x+3y). 
2k. (x 2 + 2x+4)(^ 2 -2x + 4). 

25. (3x-y)(3x + y)(x + y)(x-y). 

26. (2x + 3y)(2x-3y)(x+y)(x-y). 

27. (x 2 -xy + 3y 2 )(x 2 +xy + 3y 2 ). 

28. (x 2 -3x+5)(x 2 +3x+5). 

269.-3. (x 2 +l)(x 4 -x 2 + l). k. (a-b)(a+b)(a 2 +ab+b 2 ) 

(a 2 -a& + 6 2 ). 5. (x + 2)0r-2)O 2 +2z+4-)(x 2 -2z+4). 
6. (a-6)(a 2 + 6a+36). 7. 3(1 -3x)(l+3x + 9:r 2 ): 

8. x(x-3)(x 2 +3x + 9). 9. 2(:r+5)0 2 -5x + 25), 

10. (x-y)(x 2 +xy+y 2 )(x + y)(x 2 -xy+y 2 )(x 2 + y 2 ) 
{x 4 -x 2 y 2 -\-y i ). 11. (a + &-c)(a 2 +2a& + 6 2 + ac + &c + c 2 ). 
12. 2b(3a 2 + b 2 ). 

272.— 1. i :. 2. --.. 3. x 3 . A. 18. 5. f. 6. 1. 7. 1. S. a 2 . 

a; 2 p 6 2 

9. \. 10. 4i. 11. 1. 12. a 3 . 

a 3 2 a 

27*— 1 * 2a 3 - k — 5 £ 6 - 7 - 

ax 

8. 2xK 9. -7. 10. —,. 11. x. 12. 2yK 13. — L . 
a* xi x* 

14. ^. 15. ^-. 16. 8. 17. |. 15. f . 19. 64. 00. 8. 
«i. ^. ^. si- 23. 8. 24- i ^5. 8. 26. 5. 27. 2. 
28. f. 29q|r 4 .. 50. W- «■ ¥• **•***• 56.^/19863: 



332 MATHEMATICS FOR TECHNICAL SCHOOLS 

Page 

273.-37. ^2187. 38. #9. 39. x-y. 40. xi+xyWy-yl 
41. 8. 42. T V 43. C V U- 81. 

275.— 1. ^P, ^4* \76l 2. a^/5 1 , ^TT 2 , ^13. 3. #2*, 
V2», V2\ 4. V8», v^ $'6«. 5. V6\ 6. V30. 

7. V¥-. *. \Z4500. 9. 2^/9. 

276.— 1. 25 V7. £. 8\ZlT. 5. 14 V^ |. 5V5". 5. 16-97. 

6.12-12. 7.15-81. 8.36-74. 9.78-26. 70.31-75. 

ii. 7. _2£. 2-65. 25. -25-98. U. 7-35^ 15. 16-26. 

16.4^4. 17.6^2. 18.5^/5. 19.-9^3. £0.34-29. 

£1.96. #& 61-24, 8$. 41-57. *£ 118-78. 05.51-96. 
00. 332-55. 

278.— 1. 8-66. 0. 1-155. 5. 9-794. 4- 2-683. 5. -204. 

6. 19-596. 7. -403. 8. -045. 9. -257. 19. -315. 

11.1-702. 12.-822. 15.8-989. 14. -809. 15.1-869. 

16.2-184. 17. 101. 18.1-768. 19.1-294. 
20. 46-647. 

282.— 1. -V-, -5. 2. ¥-, -V-- 3. f, -3. 4. 2|, -1. 

5. -3, -8. 6. 3£, 1. 7. 1-72, 1-28. 8. 6-19, -807. 

9. 2-38, 4-62. 10. T % -f. 12. f, -f. 12. 3, -1. 

13. 2, |. 14. 13, f. 15. 12, -2. 16'. 3, -f. 17. 4, f 

2S.f^. &=£•=£. 20.3a,f. 

285.— 1. 5, f. 2. -i -9. 5. i 2. 4- 2, -1-72. 5. f, -5. 

6. 3, -2. 7. 1, -§. 3. ±3, ±2. 9. -&, -f. 

10. ±2, -5, -1. 11. ±3, ±4. 12. 3, -2. 13. 2\. 4. 

14. 6", 8". 15. 5". 16. 3-23 rods. 17. 3- 82", 6- 18". 
18. 4-37 sec. 19. 7-59 sec. 

287.— 1. x = 17, 11, y—11, 17. 2. x = 14, -9, y = 9, -14. 

8. x = 27, -19, y = 19, -27. 4. * = 71, 13, ?/ = 13, 71. 

5. x = ±5, y = ±7, x = ±7, y = ±5. 6. x = ±8, ±5, ?/ 
= ±5, ±8. 7. x = 13, 3, i/ = 3, 13. 8. x = ll, -8, 
y = 8, -11. 9. * = 1, 2/ = l. 10. 4", 3". 11. 5", 7". 
20. 24", 7". 



ANSWERS 333 

Page 

290.— 1. 20 tons. 2. 181-7 lb. 3. 480 ft. per sec. 
4- 126f ft. per min. 

291.— 5. f". 6. 9-97 lb. 7. 20- 2". 5. 208-9 lb. 

9. 23361+ #.P. m 10-55". 11. 2-406". 12. -285". 

29k.— 1. 11-91'. «. 11-36'. 3. 25-39 miles per hr. £ 9|. 

295.-5. 37|. 6. 30-7 ohms. 7. 2X3 3 = 1 X(3-78) 3 approx. 
5. 1-5 lb. 9. 2|'. 10. 9 sq. ft. 11. 51 -,66 cc. 

296.— 12. -236-5° C. 13. 324 ohms. 

300.— 1. $492-05. 2. $1000. 5. $1000. 4- $1150-60. 
5. $2208-05. 6. $4794-96, 6% interest. 

301.— 7. 85761-04. 8. $1128-54. 9. $3680-40, $2055-00. 

10. $7413-40. 11. $10,132-70. 12. $228-03, $683-98. 
13. Amount of premiums invested = $7325 • 60. 

301.— 1. 3*. 2. 310. 

302.— 3. 33. k- $1591-20. 5. 615^. 6. 228, 40, 60. 
7. $114-91. 8. lf$ hr. 9. 21| min. 10. 16. 

11. 3117-4 lb. 12. 80. 13. -286. 14- |*. 15. 44-4 
hr. 16. 83-3. 

303.— 17. $ll-27i 18. 5-39. 19. 4526. 20. 224^, 492. 

01. -406. 22. 38-9 sec. &f. $3-99. 2k. 62£. 

05. $122-67. 26. -65%. 07. 29-25 lb. 28. 1-04. 

09. -899. 50. 112|, 6. 
30^.— 31. $57-00. 50. 41f, 33i 25. 33. $31-99. 34. f. 

35. 28 • 28'. 36. $210 • 32. 37. 19 lb. 55. fj. 39. 8 • 67". 

40. 181-13 lb. hi. 1443-17 lb. 40. 8-6, -31, 312. 

45.280. U. 36-97 lb. 
305.— k5. 42 + . 46. 50-27 sq. ft., 76-97 sq. ft. 47. 40'. 

48. -83". 49. 12-73". 50. -942". 51. 32-78 sq. in. 

50. 77-6. 53. 11-46, 10". 54. 13-26'. 55. 5-32". 

56. -125". 57. 10-16 lb. 55. 21-5. 59. 8-38. 

306.— 60. 14-7". 61. 5-01". 60. 7i". 65. 7-96". 

64. 122140. 65. 15-75". 66. 48-26 lb. 67. 343. 

65. 1333i 640, 213i 69. $852-71. 70. $389-63. 



334 MATHEMATICS FOR TECHNICAL SCHOOLS 

Page 

307.— 71. 9-2", 2-53". 72. $42-56. 73. 4-91. 74. 8183. 

75. 235 + . 76. 118-75", 118 -06", -58. 77. $507-52. 
308.— 78. 22135. 79. 1413-72. 80. $523-93. 81. 3 -08-. 

82. 58-87. 83. 182-72. 84. 1-02 lb. 85. 19867. 

86. 587-7, 39-7 sq. ft. 87. 67-259 sq. in., 1-91 lb. 
309.— 88. 10-06, 6-31. 89. 174-13 cu. in. 90. 147-66. 

91. $241-83. 92. $109-71. 93. 1110-3. 9k- 53-31. 

95. 2575 • 7 lb. 96. 148 • 68", 147 • 92", • 52. 97. $234 • 13. 
310.— 98. $142-72. 99. 32 R.P.M. 100. 661-59 sq. in. 

311.— 3. 2", 3", 1". 4. -615. 5. 52° 30', 37° 30'. 6. 189-55 

lb. 7. 5-28". 8. 75000. 9. 4-35". 10. 72°, 72°, 36°. 

ii. 81. 12. 16", 11-314". 15. 5-03. lit. 6-48". 

15. -649. 16. 15-49". 
312—17. -5-92. IS. 36'. 19. 5-77", 11-54". £0. 1559-38. 

21. Sides 8, 12, perp. 10. 22. 23395 X10 4 . 23. 3-25 

miles. 24. -47% too great. £5. 3", 4". £6. 6". 

27. 269-66 sq. ft. 0$. 13", 15". 29. 40'. 
313.-30. 28". 91. 64'. 30. 13-27'. 3k- 6"X12". 35. Z7\. 

36. 2-92". 37. $900-00. 38. 116-52 oz. 39. 9", 12", 

15". 40. 4". 
314.— 41- 3600. ^. -0039. 



INDEX 



Abscissa, 150. 

Addendum, 209. 

Addition, in Arithmetic, 3 ; in 

Algebra, 92. 
Annulus, 121. 

Antilogarithm, 231 ; tables, 321. 
Ashlar, 54. 
Axes of reference, 150. 

B 

Bead, volume of, 257. 
Board foot, 56. 
Brackets, 78. 
Brick work, 55. 

c 

Calipers, outside, inside, herma- 
phrodite, 171 ; vernier, 177. 

Cancellation in Arithmetic, 14. 

Centring, 172 ; by hermaphrodite 
calipers, 173 ; by centre square, 
173. 

Characteristic, 229. 

Circle, 118 ; circumference, of, 
119 ; area of, 120 ; arc of, 121 ; 
sector of, 122 : segment of, 122. 

Circular ring, 121. 

Clearance, 209. 

Coefficient, 71. 

Cone, 246, 247. 

Co-ordinates, 150. 

Cylinder, 243 ; hollow, 245. 



Decimal point, 1. 
Decimal fractions, 27 ; repeat- 
ing, 28. 



Decimal equivalents, 315. 
Decorating, 68. 
Dedendum, 209. 

Division, in Arithmetic, 3, 9 ; in 
Algebra, 100 ; rule of signs, 100. 



Eaves, 61. 

Ellipse, 126 ; to construct, 127 ; 
area of, 127 ; circumference of, 
127. 

English linear measure, 37. 

Equation, simple, 84 ; simul- 
taneous, 141 ; quadratic, 279 ; 
simultaneous quadratic, 286; 
exponential, 238. 

Exponent, 74. 

Expression, Algebraic, 71. 



Factors, in Arithmetic, 14 ; in 
Algebra, 263. 

Feed, lathe cutting of, 180 ; of 
milling machine, 211. 

Field book, 112. 

Flooring, 57. 

Formulas, 103, 144. 

Fractions, definition of, 18 ; kinds 
of, 18 ; addition of, 21 ; sub- 
traction of, 21 ; multiplication 
of, 25 ; division of, 26 ; deci- 
mal, 27. * 

Frustum, 251, 252. 

G 

Gear trains, 197, 198 ; compound, 
199. 



335 



336 



MATHEMATICS FOR TECHNICAL SCHOOLS 



Gears, calculation, 207 ; reduc- 
tion in head-stock, 205 ; quick 
change, 206, 207. 

Geometrical series, 297, 298. 

Graphs, 150. 

Guage of slate, 65. 



Heel, 61. 
Head-stock, 205. 



Imaginary quantity, 284. 
Index, 74 ; laws, 75, 270. 
Index plate, 213, 215, 219. 
Indexing, rapid, 213 ; plain, 214 

differential, 217. 
Irrational quantity, 274. 
Irregular figures, area of, ]29. 



Lathe, cutting speed of, 179; 

compound geared, 202; 203; 

lead of, 201 ; simple geared 

200. 
Lathing, 57. 
Lead screw, 200. 
Logarithm, 228, 229 ; of number 

less than unity, 235; of a 

power, 236; tables, 319. 
Lumber, 56. 

M 

Machinist's scale, 171. 

Mantissa, 229. • 

•Measure, linear, English, 37; 
linear metric, 38 ; square, 
English, 40 ; square, metric, 
40; cubic, English, 42; pubic, 
metric, 42. 

Micrometer, 175. 



Milling machine, 210; cutting 
speed of, 210; feed of, 211, 212 ; 
lead of, 223; change gear cal- 
culation, 223. 

Multiple, least common, 22. 

Multiplication, in Arithmetic, 3, 
5; tables, 6; in Algebra, 94'; 
rule of signs, 95. 

N 
Negative quantities, 81. 
Notation, in Algebra, 71 ; i n 
Arithmetic, 1. 



Ordinate, 150. 

Origin, 150. 

Outside diameter, 209. 



Painting, 68, 69. 

Parallelogram, 109. 

Percentage, 30. 

Pictograph, 162. 

Pitch, of roof, 61. 

Pitch, diameter, 208 ; circle, 208 ; 

diametral, 208 ; circular, 208. 
Planimeter, 130, 131, 132. 
Plate, 61. 
Plastering, 67. 
Polygon, area of, 127. 
Power of, 10, 7 ; ofaquantity, 71. 
Present worth. 299. 
Prism, 241, 212. 
Prismoid, 259. 
Proportion, 135; inverse, 138; in 

similar triangles, 136. 
Pyramid, 249. 
n, value of, lift, 

Q 

Quadratic equations, 286. 



INDEX 



337 



R 

Rafters, 01; hip, 62, 63; jack, 

62, 63. 
Ratio, 134. 

Rational quantity, 274. 
Rectangle, 108. 
Ring, solid, 258 ; anchor, 258. 
Rise, 61. 

Roofs, gable, 61 ; hip, cottage, 62. 
Roofing, 64. 

Root, square, 53 ; in Algebra, 76. 
Run, 61. 
Rubble, 54. 



Screw, 188. 

Shingles, 64. 

Signs of operation in Arith- 
metic, 13. 

Simpson's Rule, 129. 

Simultaneous equations, 141 ; 
simultaneous quadratics, 286. 

Slate, 64. 

Span, 61. 

Specific gravity, 46 ; tables of, 
316, 317, 318.* 

Sphere, 254, 255 : sector of, 257 ; 
segment of, 256 ; zone of, 257. 

Spirals, cutting, 221 ; position of 
table, 224. 

Square, 108. 

Square root, 50. 

Stone work, table, 54. 

Subtraction, in Arithmetic, 3, 5 ; 
in Algebra, 93. 

Surds, 273; quadratic, 274; like 
and unlike, 274 ; addition of, 
275 ; subtraction of, 275 ; mul- 
tiplication of, 275 ; mixed and 
entire, 276 ; division of, 277. 



Symbols, of Arithmetic, 1 ; of 
Algebra, 71. 



Taper, as amount, 183 ; as angle, 
183; Morse, 183; B. & S„ 184; 
Jarno, 184 ; cutting by com- 
pound rest, 184, 185 ; cutting 
by offsetting tailstock, 184, 185, 
186 ; cutting by taper attach- 
ment, 186. 

Terms, like and unlike, 77. 

Threads, pitch, 188; diameter of, 
188, 209 ; inside diameter of, 
188, 209 ; single, double, triple, 
189 ; right-handed, left-handed, 
189 ; double, triple cutting, 205 ; 
sharp " V," 190 ; U.S. Std., 192 ; 
Square, 193; Acme 29°, 194; 
Whitworth, 196. 

Thread cutting, 197, 200. 

Toise, 54. 

Trapezium, 111. 

Triangle, 109, 110. 

Trigonometrical ratios, 182. 

Try square, 171. 



Variation, 288. 
Vernier, 173, 174. 
Vernier caliper, 177. 



W 
AVedge, 258. 
Whole depth, 208. 
Working depth, 208.